Question 13 Marks
Solve the following quadratic equation:
$3\text{x}^2-2\sqrt6\text{x}+2=0$
Answer$3\text{x}^2-2\sqrt6\text{x}+2=0$
$\Rightarrow3\text{x}^2-\sqrt6\text{x}-\sqrt6\text{x}+2=0$
$\Rightarrow\sqrt3\text{x}\big(\sqrt3\text{x}-\sqrt2\big)-\sqrt2\big(\sqrt3\text{x}-\sqrt2\big)=0$
$\Rightarrow\big(\sqrt3\text{x}-\sqrt2\big)\big(\sqrt3\text{x}-\sqrt2\big)=0$
$\Rightarrow\big(\sqrt3\text{x}-\sqrt2\big)^2=0$
$\Rightarrow\sqrt3\text{x}-\sqrt2=0$
$\Rightarrow\text{x}=\frac{\sqrt2}{\sqrt3}$ (repeated root)
View full question & answer→Question 23 Marks
Solve the following quadratic equation:
$ x^2+5 x-\left(a^2+a-6\right)=0 $
Answer$ x^2+5 x-\left(a^2+a-6\right)=0 $
$ \Rightarrow x^2+5 x-\left(a^2+3 a-2 a-6\right)=0 $
$ \Rightarrow x^2+5 x-[a(a+3)-2(a+3)]=0 $
$ \Rightarrow x^2+5 x-[(a+3)(a-2)]=0 $
$ \Rightarrow x^2+(a+3) x-(a-2) x-(a+3)(a-2)=0 $
$ \Rightarrow x[x+(a+3)]-(a+2)[x+(a+3)]=0 $
$ \Rightarrow[x+(a+3)][x-(a-2)]=0 $
$ \Rightarrow x+(a+3)=0 \text { or } x-(a-2)=0 $
$ \Rightarrow x=-(a+3) \text { or } x=a-2$
View full question & answer→Question 33 Marks
Solve the following quadratic equation:
$ x^2-(2 b-1) x+\left(b^2-b-20\right)=0 $
Answer$ x^2-(2 b-1) x+\left(b^2-b-20\right)=0 $
$ \Rightarrow x^2-(2 b-1) x+\left(b^2-5 b+4 b-20\right)=0 $
$ \Rightarrow x^2-(2 b-1) x+[b(b-5)+4(b-5)]=0 $
$ \Rightarrow x^2-(2 b-1) x+(b-5)(b+4)=0 $
$ \Rightarrow x^2-(2 b-1) x-(b+4) x+(b-5)(b+4)=0 $
$ \Rightarrow x[x-(b-5)]-(b+4)[x-(b-5)]=0 $
$ \Rightarrow[x-(b-5)][x-(b+4)]=0 $
$ \Rightarrow x-(b-5)=0 \text { or } x-(b+4)=0 $
$ \Rightarrow x=b-5 \text { or } x=b+4$
View full question & answer→Question 43 Marks
Solve the following quadratic equation:
$ 2 x^2+a x-a^2=0 $
Answer$ 2 x^2+a x-a^2=0 $
$ \Rightarrow 2 x^2+2 a x-a x-a^2+2=0 $
$ \Rightarrow 2 x(x+a)-a(x+a)=0 $
$ \Rightarrow(x+a)(2 x-a)=0 $
$ \Rightarrow x+a=0 \text { or } 2 x-a=0 $
$ \Rightarrow x=-a, x=\frac{a}{2}$
View full question & answer→Question 53 Marks
If $-4$ is a root of the quadratic equation $x^2+2 x+4 p=0$, find the value of $k$ for which the quadratic equation $x^2+ px(1 + 3k) + 7(3 + 2k) = 0$ has equal roots.
AnswerIt is given that $-4$ is a root of the quadratic equation $x^2+2 x+4 p=0$.
$ \therefore(-4)^2+2 \times(-4)+4 p=0 $
$ \Rightarrow 16-8+4 p=0 $
$ \Rightarrow 4 p+8=0 $
$ \Rightarrow p=-2$
The equation $\mathrm{x}^2+\mathrm{px}(1+3 \mathrm{k})+7(3+2 \mathrm{k})=0$ has equal roots.
$ \therefore D=0 $
$ \Rightarrow[p(1+3 k)]^2-4 \times 1 \times 7(3+2 k)=0 $
$ \Rightarrow-[2(1+3 k)]^2-28(3+2 k)=0 $
$ \Rightarrow 4\left(1+6 k+9 k^2\right)-28(3+2 k)=0 $
$ \Rightarrow 4\left(1+6 k+9 k^2-21-14 k\right)=0 $
$ \Rightarrow 9 k^2-8 k-20=0 $
$ \Rightarrow 9 k^2-18 k+10 k-20=0 $
$ \Rightarrow 9 k(k-2)+10(k-2)=0 $
$ \Rightarrow(k-2)(9 k+10)=0 $
$ \Rightarrow k-2=0 \text { or } 9 k+10=0 $
$ \Rightarrow k=2 \text { or } k=-\frac{10}{9}$
Hence, the required value of $k$ is $2$ or $-\frac{10}{9}$.
View full question & answer→Question 63 Marks
Solve the following quadratic equation:
$\text{10x}-\frac{1}{\text{x}}=3$
Answer$\text{10x}-\frac{1}{\text{x}}=3$
$ \Rightarrow 10 x^2-1=3 x $
$ \Rightarrow 10 x^2-5 x+2 x+1=0 $
$ \Rightarrow 5 x(2 x-1)+1(2 x-1)=0 $
$ \Rightarrow(5 x+1)(2 x-1)=0 $
$ \Rightarrow(5 x+1)(2 x-1)=0 $
$ \Rightarrow 5 x+1=0 \text { or } 2 x-1=0$
$\Rightarrow\text{x}=\frac{-1}{5}$ or $\text{x}=\frac{1}{2}$
Hence, $\frac{-1}{5}$ and $\frac{1}{2}$ are the roots of the given equation.
View full question & answer→Question 73 Marks
Solve the following quadratic equation:
$ 3 x^2-2 x-1=0 $
Answer$ 3 x^2-2 x-1=0 $
$ \Rightarrow 3 x^2-3 x+1 x-1=0 $
$ \Rightarrow 3 x(x-1)+1(x-1)=0 $
$ \Rightarrow(x-1)(3 x+1)=0 $
$ \Rightarrow x-1=0 \text { or } 3 x+1=0 $
$ \Rightarrow x=1 \text { or } x=\frac{-1}{3}$
$\text { Hence, } 1 \text { and } \frac{-1}{3} \text { are the roots of the equation } 3 x^2-2 x-1=0$
View full question & answer→Question 83 Marks
Solve the following equations by using the method of completing the square:
$ x^2-4 x+1=0 $
Answer$ x^2-4 x+1=0 $
$ \Rightarrow x^2-4 x=-1 $
$ \Rightarrow x^2-2 \times x \times 2+2^2=-1+2^2 \text { (Adding } 2^2 \text { on both sides) } $
$ \Rightarrow(x-2)^2=-1+4=3$
$\Rightarrow\text{x}-2=\pm\sqrt3$ (Taking square root on both sides)
$\Rightarrow\text{x}-2=\sqrt3$ or $\text{x}-2=-\sqrt3$
$\Rightarrow\text{x}=2+\sqrt3$ or $\text{x}=2-\sqrt3$
Hence, $2+\sqrt3$ and $2-\sqrt3$ are the roots of the given equation.
View full question & answer→Question 93 Marks
Determine the values of p for which the quadratic equation $2 x^2+p x+8=0$ has real roots.
AnswerGiven:
$2 x^2+p x+8=0$
Here,
$\mathrm{a}=2, \mathrm{~b}=\mathrm{p} \text { and } \mathrm{c}=8$
Discriminant $D$ is given by:
$ D=\left(b^2-4 a c\right)$
$ D=p^2-4 \times 2 \times 8 $
$D=\left(p^2-64\right)$
If $\text{D}\ge0,$ the roots of the equation will be real.
$\Rightarrow(\text{p}^2-64)\ge0$
$\Rightarrow(\text{p}+8)(\text{p}-8)\ge0$
$\Rightarrow\text{p}\ge8$ and $\text{p}\le-8$
Thus, the roots of the equation are real for $\text{p}\ge8$ and $\text{p}\le-8.$
View full question & answer→Question 103 Marks
Solve the following equations by using the method of completing the square:
$ 8 x^2-14 x-15=0 $
Answer$ 8 x^2-14 x-15=0 $
$ \Rightarrow 16 x^2-28 x-30=0(\text { Multiplying both sides by } 2)$
$ \Rightarrow 16 x^2-28 x=30$
$\Rightarrow(\text{4x})^2-2\times\text{4x}\times\frac{7}{2}+\Big(\frac{7}{2}\Big)^2\\=30+\Big(\frac{7}{2}\Big)^2$ $\Big[$Adding $\Big(\frac{7}2{}\Big)^2$ on both sides$\Big]$
$\Rightarrow\Big(\text{4x}-\frac{7}{2}\Big)^2$
$=30+\frac{49}{4}$
$=\frac{169}{4}=\Big(\frac{13}{2}\Big)^2$
$\Rightarrow\text{4x}-\frac{7}{2}=\pm\frac{13}{2}$ (Taking square root on both sides)
$\Rightarrow\text{4x}-\frac{7}{2}=\frac{13}{2}$ or $\text{4x}-\frac{7}{2}=-\frac{13}{2}$
$\Rightarrow\text{4x}=\frac{13}{2}+\frac{7}{2}=\frac{20}{2}=10$ or $\text{4x}=-\frac{13}{2}+\frac{7}{2}=-\frac{6}{2}=-3$
$\Rightarrow\text{x}=\frac{5}{2}$ or $\text{x}=-\frac{3}4{}$
Hence, $\frac{5}{2}$ and $-\frac{3}{4}$ are the roots of the given equation.
View full question & answer→Question 113 Marks
Find the values of k for which the given quadratic equation has real and distinct roots:
$k x^2+6 x+1=0$
Answer$\text { The given equation is } k x^2+6 x+1=0$
$\therefore D=6^2-4 \times k \times 1$
$D = 36 - 4k$
The given equation has real and distinct roots if $D > 0.$
$\therefore 36 - 4k > 0$
$⇒ 4k < 36$
$⇒ k < 9$
View full question & answer→Question 123 Marks
Solve the following quadratic equation:
$\text{2x}^2-\text{x}+\frac{1}{8}=0$
Answer$\text{2x}^2-\text{x}+\frac{1}{8}=0$
$ \Rightarrow 16 x^2-8 x+1=0$
$ \Rightarrow 16 x^2-4 x-4 x+1=0$
$ \Rightarrow 4 x(4 x-1)-1(4 x-1)=0$
$ \Rightarrow(4 x-1)(4 x-1)=0$
$ \Rightarrow(4 x-1)^2=0$
$ \Rightarrow 4 x-1=0$
$\Rightarrow\text{x}=\frac{1}{4}$ (repeated root)
View full question & answer→Question 133 Marks
If the roots of the equation $\left(c^2-a b\right) x^2-2\left(a^2-b c\right) x+\left(b^2-a c\right)=0$ are real and equal, show that either $a = 0$ or $(a^3+ b^3+ c^3) = 3abc.$
AnswerGiven:
$\left(c^2-a b\right) x^2-2\left(a^2-b c\right) x+\left(b^2-a c\right)=0$
Here,
$a=\left(c^2-a b\right), b=-2\left(a^2-b c\right), c=\left(b^2-a c\right)$
It is given that the roots of the equation are real and equal; therefore, we have:
$D=0 $
$\Rightarrow\left(b^2-4 a c\right)=0 $
$\Rightarrow\left\{-2\left(a^2-b c\right)\right\}^2-4 \times\left(c^2-a b\right) \times\left(b^2-a c\right)=0 $
$\Rightarrow 4\left(a^4-2 a^2 b c+b^2 c^2\right)-4\left(b^2 c^2-a c^3-a b^3+a^2 b c\right)=0 $
$\Rightarrow a^4-2 a^2 b c+b^2 c^2-b^2 c^2+a c^3+a b^3-a^2 b c=0 $
$\Rightarrow a^4-3 a^2 b c+a c^3+a b^3=0 $
$\Rightarrow a\left(a^3-3 a b c+c^3+b^3\right)=0$
Now,
$ a=0 \text { or } a^3-3 a b c+c^3+b^3=0 $
$a=0 \text { or } a^3+b^3+c^3=3 a b c$
View full question & answer→Question 143 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:
$36 x^2-12 a x+\left(a^2-b^2\right)=0$
AnswerThe given equation is $36 x^2-12 a x+\left(a^2-b^2\right)=0$.
Comparing it with $A x^2+B x+C=0$, we get
$A=36, B=-12 a$ and $C=a^2-b^2$
$ \therefore \text { Discriminant, } D=B^2-4 A C $
$=(-12 a)^2-4 \times 36 \times\left(a^2-b^2\right) $
$=144 a^2-144 a^2+144 b^2$
$=144b^2 > 0$
So, the given equation has real roots.
Now, $\sqrt{\text{D}}=\sqrt{144\text{b}^2}=12\text{b}$
$\therefore\alpha=\frac{-\text{B}+\sqrt{\text{D}}}{2\text{A}}$
$=\frac{-(-12\text{a}+12\text{b})}{2\times36}$
$=\frac{12(\text{a}+\text{b})}{72}$
$=\frac{\text{a}+\text{b}}{6}$
$\beta=\frac{-\text{B}-\sqrt{\text{D}}}{2\text{A}}$
$=\frac{-(-12\text{a})-12\text{b}}{2\times36}$
$=\frac{12(\text{a}-\text{b})}{72}$
$=\frac{\text{a}-\text{b}}{6}$
Hence, $\frac{\text{a}+\text{b}}{6}$ and $\frac{\text{a}-\text{b}}{6}$ are the roots of the given equation.
View full question & answer→Question 153 Marks
The following are the roots of $3x^2+ 2x - 1 = 0?$
$-\frac{1}{2}$
AnswerThe given equation is $3x^2+ 2x - 1 = 0$
On substituting $\text{x}=-\frac{1}{2}$ in the equation, we get
$\text{LHS}=3\times\Big(\frac{-1}{2}\Big)^2+2\times\Big(\frac{-1}{2}\Big)-1=0$
$=\frac{3}{4}-1+1\neq0$
$\therefore\ \text{LHS}\neq\text{RHS}$
$\therefore\ \text{x}=\frac{-1}{2}$ is not a solution of $3x^2+ 2x - 1 = 0$
View full question & answer→Question 163 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:
$2 x^2+x-4=0$
AnswerGiven,
$2 x^2+x-4=0$
On comparing it with $a x^2+b x+c=0$,
we get $a=2, b=1$ and $c=-4$
Discriminant $D$ is given by:
$ D=\left(b^2-4 a c\right) $
$=(1)^2-4 \times 2 \times(-4)$
$= 1 + 32$
$= 33 > 0$
So, the given equation has real roots.
Now, $\sqrt{\text{D}}=\sqrt{33}$
$\therefore\ \alpha=\frac{-\text{b}+\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-1+\sqrt{33}}{2\times2}$
$=\frac{-1+\sqrt{33}}{4}$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-1-\sqrt{33}}{2\times2}$
$=\frac{-1-\sqrt{33}}{4}$
Hence $\frac{-1+\sqrt{33}}{4}$ and $\frac{-1-\sqrt{33}}{4}$ are the roots of the given equation.
View full question & answer→Question 173 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:
$4\sqrt3\text{x}^2+5\text{x}-2\sqrt3=0$
AnswerThe given equation is:
$4\sqrt3{\text{x}}^2+5\text{x}-2\sqrt3=0$
Comparing it with $ax^2+ bx + c = 0,$ we get
$\text{a}=4\sqrt3,\ \text{b}=5$ and $\text{c}=-2\sqrt3$
$\therefore$ Discriminant, $\text{D}=\text{b}^2-4\text{ac}$
$=5^2-4\times4\sqrt3\times\big(-2\sqrt3\big)$
$=25+96$
$=121>0$
So, the given equation has real roots.
Now, $\sqrt{\text{D}}=\sqrt{121}=11$
$\therefore\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-5+11}{2\times4\sqrt3}$
$=\frac{6}{8\sqrt3}$
$=\frac{\sqrt3}{4}$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-5-11}{2\times4\sqrt3}$
$=\frac{-16}{8\sqrt3}$
$=-\frac{2\sqrt3}{3}$
Hence, $\frac{\sqrt3}{4}$ and $-\frac{2\sqrt3}{3}$ are the roots of the given equation.
View full question & answer→Question 183 Marks
Solve the following equations by using the method of completing the square:
$5 x^2-6 x-2=0$
Answer$5 x^2-6 x-2=0$
$\left.\Rightarrow 25 x^2-30 x-10=0 \text { (Multiplying both sides by } 5\right)$
$\Rightarrow 25 x^2-30 x=10$
$\Rightarrow(5 x)^2-2 \times 5 x \times 3+3^2=10+3^2 \text { [Adding } 3^2 \text { on both sides] }$
$\Rightarrow(5 x-3)^2=10+9=19$
$\Rightarrow\text{5x}-3=\pm19$ (Taking square root on both sides)
$\Rightarrow\text{5x}-3=\sqrt{19}$ or $\text{5x}- 3=-\sqrt{19}$
$\Rightarrow\text{5x}=3+\sqrt{19}$ or $\text{5x}=3-\sqrt{19}$
$\Rightarrow\text{x}=\frac{3+\sqrt{19}}{5}$ or $\text{x}=\frac{3-\sqrt{19}}{5}$
Hence, $\frac{3+\sqrt{19}}{5}$ and $\frac{3-\sqrt{19}}{5}$ are the roots of the given equation.
View full question & answer→Question 193 Marks
Find the non-zero value of k for which the roots of the quadratic equation $9x^2- 3kx + k = 0$ are real and equal.
AnswerThe given equation is $9x^2- 3kx + k = 0.$
This is of the form $ax^2+ bx + c = 0$, where $a = 9, b = -3k$ and $c = k.$
$\therefore D = b^2- 4ac$
$D = (-3k)^2- 4 × 9 × k$
$D = 9k^2- 36k$
The given equation will have real and equal roots if $D = 0.$
$\therefore 9k^2- 36k = 0$
$⇒ 9k(k - 4) = 0$
$⇒ k = 0$ or $ k - 4 = 0$
$⇒ k = 0$ or $k = 4$
But, $k ≠ 0$ (Given)
Hence, the required value of $k$ is $4.$
View full question & answer→Question 203 Marks
Solve the following quadratic equation:
$\sqrt2\text{x}^2+7\text{x}+5\sqrt2=0$
Answer$\sqrt2\text{x}^2+7\text{x}+5\sqrt2=0$
$\Rightarrow\sqrt2\text{x}^2+5\text{x}+2\text{x}+50\sqrt2=0$
$\Rightarrow\text{x}\big(\sqrt2\text{x}+5\big)-\sqrt2\big(\sqrt2\text{x}+5\big)=0$
$\Rightarrow\big(\sqrt2\text{x}+5\big)\big(\text{x}+\sqrt2\big)=0$
$\Rightarrow\sqrt2\text{x}+5=0$ or $\text{x}+\sqrt2=0$
$\Rightarrow\text{x}=\frac{-5}{\sqrt2}$ or $\text{x}=-\sqrt2$
View full question & answer→Question 213 Marks
Solve the following quadratic equation:
$\sqrt3\text{x}^2+\text{10x}+7\sqrt3=0$
Answer$\sqrt3\text{x}^2+\text{10x}+7\sqrt3=0$
$\Rightarrow\sqrt3\text{x}^2+12\text{x}-2\text{x}-8\sqrt3=0$
$\Rightarrow\sqrt3\text{x}\big(\text{x}+4\sqrt3\big)-2\big(\text{x}+4\sqrt{3}\big)=0$
$\Rightarrow\big(\sqrt{3}\text{x}-2\big)\big(\text{x}+4\sqrt3\big)=0$
$\Rightarrow\sqrt3\text{x}-2=0$ or $\text{x}+4\sqrt3=0$
$\Rightarrow\text{x}=\frac{2}{\sqrt3}$ or $\text{x}=-4\sqrt3$
View full question & answer→Question 223 Marks
Solve the following quadratic equation:
$4\sqrt6\text{x}^2-\text{13x}-2\sqrt6=0$
Answer$4\sqrt6\text{x}^2-\text{13x}-2\sqrt6=0$
$\Rightarrow4\sqrt6\text{x}^2-16\text{x}+3\text{x}-2\sqrt6=0$
$\Rightarrow4\sqrt2\text{x}\big(\sqrt3\text{x}-2\sqrt2\big)+\sqrt3\big(\sqrt3\text{x}-2\sqrt2\big)=0$
$\Rightarrow\big(\sqrt3\text{x}-2\sqrt2\big)\big(4\sqrt2\text{x}+\sqrt3\big)=0$
$\Rightarrow\Big(\frac{2\sqrt2}{\sqrt3}\times\frac{\sqrt3}{\sqrt3}\Big)=\frac{2\sqrt6}{3}$ or $\Big(\frac{-\sqrt3}{4\sqrt2}\times\frac{\sqrt2}{\sqrt2}\Big)=\frac{-\sqrt6}{8}$
Hence, $\frac{2\sqrt6}{3}$ and $\frac{-\sqrt6}{8}$ are the roots of the given equation.
View full question & answer→Question 233 Marks
If $-5$ is a root of the quadratic equation $2 x^2+p x-15=0$ and the quadratic equation $p(x^2+ x) + k = 0$ has equal roots, find the value of $k$.
AnswerIt is given that $-5$ is a root of the quadratic equation $2 x^2+p x-15=0$.
$\therefore 2(-5)^2+p \times(-5)-15=0$
$⇒ -5p + 35 = 0$
$⇒ p = 7$
The roots of the equation $p x^2+p x+k=0$ are equal.
$\therefore D = 0$
$\Rightarrow p^2-4 p k=0$
$ \Rightarrow(7)^2-4 \times 7 \times \mathrm{k}=0$
$⇒ 49 - 28k = 0$
$\Rightarrow\text{k}=\frac{49}{28}=\frac{7}{4}$
Thus, the value of $k$ is $\frac{7}{4}.$
View full question & answer→Question 243 Marks
Solve the following quadratic equation:
$\text{x}^2-3\sqrt5\text{x}+10=0$
Answer$\text{x}^2-3\sqrt5\text{x}+10=0$
$\Rightarrow\text{x}^2-\sqrt5\text{x}-2\sqrt{5}\text{x}+10=0$
$\Rightarrow\text{x}\big(\text{x}-\sqrt5\big)-2\sqrt5\big(\text{x}-\sqrt5\big)=0$
$\Rightarrow\big(\text{x}-\sqrt5\big)\big(\text{x}-2\sqrt5\big)=0$
$\Rightarrow\text{x}-\sqrt5=0$ or $\text{x}-2\sqrt5=0$
$\Rightarrow\text{x}=\sqrt{5}$ or $\text{x}=2\sqrt5$
View full question & answer→Question 253 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:
$\sqrt3\text{x}^2+10\text{x}-8\sqrt3=0$
AnswerGiven: $\sqrt3\text{x}^2+10\text{x}-8\sqrt3=0$
On comparing it with $ax^2+ bx + c = 0,$ we get:
$\text{x}=\sqrt3,\ \text{b}=10$ and $\text{c}=-8\sqrt3$
Discriminant D is given by:
$D = (b^2- 4ac)$
$=(10)^2-4\times\sqrt3\times\big(8\sqrt3\big)$
$=100+96$
$=196>0$
Hence, the roots of the equation are real.
Roots $\alpha$ and $\beta$ are given by:
$\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{\text{2a}}$
$=\frac{-10+\sqrt{196}}{2\sqrt3}$
$=\frac{-10+14}{2\sqrt3}$
$=\frac{4}{2\sqrt3}$
$=\frac{2}{\sqrt3}$
$=\frac{2}{\sqrt3}\times\frac{\sqrt3}{\sqrt3}$
$=\frac{2\sqrt3}{3}$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{\text{2a}}$
$=\frac{-10-\sqrt{196}}{2\sqrt3}$
$=\frac{-10-14}{2\sqrt3}$
$=\frac{-24}{2\sqrt3}$
$=\frac{-12}{\sqrt3}$
$=\frac{-12}{\sqrt3}\times\frac{\sqrt3}{\sqrt3}$
$=\frac{-12\sqrt3}{3}$
$=-4\sqrt3$
Thus, the roots of the equation are $\frac{2\sqrt3}{3}$ and $-4\sqrt3.$
View full question & answer→Question 263 Marks
Solve the following quadratic equation:
$5 x^2+13 x+8=0$
Answer$5 x^2+13 x+8=0$
$\Rightarrow 5 x^2+5 x+8 x+8=0$
$\Rightarrow 5 x(x+1)+8(x+1)=0$
$\Rightarrow(x+1)(5 x+8)=0$
$\Rightarrow x+1=0 \text { or } 5 x+8=0$
$\Rightarrow x=-1 \text { or } x=\frac{-8}{5}$
View full question & answer→Question 273 Marks
If the equation $\left(1+m^2\right) x^2+2 m c x+\left(c^2-a^2\right)=0$ has equal roots, prove that $c^2=a^2\left(1+m^2\right)$.
AnswerGiven:
$(1 + m^2)x^2+ 2mcx + (c^2- a^2) = 0$
Here,
$a = (1 + m^2), b = 2mc$ and $c = (c^2- a^2)$
It is given that the roots of the equation are equal; therefore, we have:
$ D=0$
$\Rightarrow\left(b^2-4 a c\right)=0$
$\Rightarrow(2 m c)^2-4 \times\left(1+m^2\right) \times\left(c^2-a^2\right)=0$
$\Rightarrow 4 m^2 c^2-4\left(c^2-a^2+m^2 c^2-m^2 a^2\right)=0$
$\Rightarrow 4 m^2 c^2-4 c^2+4 a^2-4 m^2 c^2+4 m^2 a^2=0$
$\Rightarrow-4 c^2+4 a^2+4 m^2 a^2=0$
$\Rightarrow a^2+m^2 a^2=c^2$
$\Rightarrow a^2\left(1+m^2\right)=c^2$
$\Rightarrow c^2=a^2\left(1+m^2\right)$
Hence proved.
View full question & answer→Question 283 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:
$x^2-6 x+4=0$
AnswerGiven,
$x^2-6 x+4=0$
$\text { On comparing it with } a x^2+b x+c=0 \text {, we get }$
Discriminant D is given by:
$a = 1, b = -6$ and $c = 4$
$D = (b^2- 4ac)$
$= (-6)^2- 4 × 1 × 4$
$= 36 - 16$
$= 20 > 0$
Hence, the roots of the equation are real.
Roots $\alpha$ and $\beta$ are given by,
$\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-(-6)+\sqrt{20}}{2\times1}$
$=\frac{6+2\sqrt5}{2}$
$=\frac{2\big(3+\sqrt5\big)}{2}$
$=\big(3+\sqrt5\big)$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-(-6)-\sqrt{20}}{2\times1}$
$=\frac{6-2\sqrt5}{2}$
$=\frac{2\big(3-\sqrt5\big)}{2}$
$=\big(3-\sqrt5\big)$
Thus, the roots of the equation are $\big(3+2\sqrt5\big)$ and $\big(3-2\sqrt5\big).$
View full question & answer→Question 293 Marks
Solve the following equations by using the method of completing the square:
$3 x^2-x-2=0 $
Answer$3 x^2-x-2=0 $
$ \Rightarrow 9 x^2-3 x-6=0 \text { (Multiplying both sides by } 3 \text { ) } $
$ \Rightarrow 9 x^2-3 x=6$
$\Rightarrow(\text{3x})^2-2\times\text{3x}\times\frac{1}{2}+\Big(\frac{1}{2}\Big)^2\\=6+\Big(\frac{1}{2}\Big)^2$ $\Big[$Adding $\Big(\frac{1}2{}\Big)^2$ on both sides$\Big]$
$\Rightarrow\Big(\text{3x}-\frac{1}{2}\Big)^2$
$=6+\frac{1}{4}$
$=\frac{25}{4}=\Big(\frac{5}{2}\Big)^2$
$\Rightarrow\text{3x}-\frac{1}{2}=\pm\frac{5}{2}$ (Taking square root on both sides)
$\Rightarrow\text{3x}-\frac{1}{2}=\frac{5}{2}$ or $\text{3x}-\frac{1}{2}=-\frac{5}{2}$
$\Rightarrow\text{3x}=\frac{5}{2}+\frac{1}{2}=\frac{6}{2}=3$ or $\text{3x}=-\frac{5}{2}+\frac{1}{2}=-\frac{4}{2}=-2$
⇒ x = 1 or $\text{x}=-\frac{2}3{}$
Hence 1 and $-\frac{2}{3}$ are the roots of the given equation.
View full question & answer→Question 303 Marks
For what values of k are the roots of the quadratic equation $3 \mathrm{x}^2+2 \mathrm{kx}+27=0$ real and equal?
AnswerGiven:
$3 \mathrm{x}^2+2 \mathrm{kx}+27=0$
Here,
$a=3, b=2 k \text { and } c=27$
It is given that the roots of the equation are real and equal; therefore, we have:
$ D=0 $
$ \Rightarrow(2 k)^2-4 \times 3 \times 27=0 $
$ \Rightarrow 4 k^2-324=0 $
$ \Rightarrow 4 k^2=324 $
$ \Rightarrow k^2=81 $
$ \Rightarrow k= \pm 9 $
$ \therefore k=9 \text { or } k=-9$
View full question & answer→Question 313 Marks
Find the value of $k$ for which $x = 1$ is a root of the equation $x^2+ kx + 3 = 0$ Also, find the other root.
AnswerSince $x = 1$ is a solution of $x^2+ kx + 3 = 0$, it must be satisfy the equation.
$(1)^2+ k(1) + 3 = 0$
$⇒ k = -4$
Hence the required value of $k = -4$
Since $\text{x}=\frac{3}{4}$ is a root of $ax^2+ bx - 6 = 0$, we have $\text{a}\times\Big(\frac{3}{4}\Big)^2+\text{b}\times\Big(\frac{3}{4}\Big)-6=0$
$\Rightarrow\frac{\text{9a}}{16}+\frac{\text{3b}}{4}-6=0$
$⇒ 9a + 12b = 96$
$⇒ 3a + 4b = 32 ...(1)$
Again $x = -2$ being a root of $ax^2+ bx - 6 = 0,$ we have
$a × (-2) + b(-2) - 6 = 0$
$⇒ 4a - 2b = 6$
$⇒ 2a - b = 3 ...(2)$
View full question & answer→Question 323 Marks
Solve the following equations by using the method of completing the square:
$ 4 x^2+4 b x-\left(a^2-b^2\right)=0 $
Answer$ 4 x^2+4 b x-\left(a^2-b^2\right)=0 $
$ \Rightarrow 4 x^2+4 b x=a^2-b^2 $
$ \Rightarrow(2 x)^2+2 \times 2 x \times b+b^2=a^2-b^2+b^2 \text { [Adding } b^2 \text { on both sides] } $
$ \Rightarrow(2 x+b)^2=a^2$
$\Rightarrow\text{2x}+\text{b}=\pm\text{a}$ (Taking square root on both sides)
$⇒ 2x + b = a$ or $2x + b = -a$
$⇒ 2x = a - b$ or $2x = -a - b$
$\Rightarrow\text{x}=\frac{\text{a}-\text{b}}{2}$ or $\text{x}=-\frac{\text{a}+\text{b}}{2}$
Hence, $\frac{\text{a}-\text{b}}{2}$ and $-\frac{\text{a}+\text{b}}{2}$ are the roots of the given equation.
View full question & answer→Question 333 Marks
Solve the following equations by using the method of completing the square:
$ x^2-6 x+3=0 $
Answer$ x^2-6 x+3=0 $
$ \Rightarrow x^2-6 x=-3 $
$ \Rightarrow x^2-2 \times x \times 3+3^2=-3+3^2 \text { (Adding } 3^2 \text { on both sides) } $
$ \Rightarrow(x-3)^2=-3+9=6$
$\Rightarrow\text{x}-3=\pm\sqrt6$ (Taking square root on both sides)
$\Rightarrow\text{x}-3=\sqrt6$ or $\text{x}-3=-\sqrt6$
$\Rightarrow\text{x}=3+\sqrt6$ or $\text{x}=3-\sqrt6$
Hence, $3+\sqrt6$ and $3-\sqrt6$ are the roots of the given equation.
View full question & answer→Question 343 Marks
Solve the following quadratic equation:
$ 4 x^2+4 b x-\left(a^2-b^2\right)=0 $
Answer$ 4 x^2+4 b x-\left(a^2-b^2\right)=0 $
$ \Rightarrow 4 x^2+4 b x+\left(b^2-a^2\right)=0 $
$ \Rightarrow 4 x^2+2(b+a) x+2(b-a) x+\left(b^2-a^2\right)=0 $
$ \Rightarrow 2 x[2 x+(b+a)]+(b-a)[2 x+(b+a)]=0 $
$ \Rightarrow[2 x+(b+a)][2 x+(b-a)]=0 $
$ \Rightarrow 2 x+(b-a)=0 \text { or } 2 x+9 b-a=0$
$\Rightarrow\text{x}=\frac{-(\text{b}+\text{a})}{2}$ or $\text{x}=\frac{-(\text{b}-\text{a})}{2}$
$\Rightarrow\text{x}=\frac{-(\text{a}+\text{b})}{2}$ or $\text{x}=\frac{(\text{a}-\text{b})}{2}$
View full question & answer→Question 353 Marks
The following are quadratic equations in $x?$
$\text{x}^2-\frac{1}{\text{x}^2}=5$
Answer$\text{x}^2-\frac{1}{\text{x}^2}=5$
$ \Rightarrow x^4-1=5 x^2 $
$\Rightarrow x^4-5 x^2-1=0$
$\text { And }\left(x^4-5 x^2-1\right) \text { Being a polynomial of degree } 4$
$\therefore\ \text{x}^2-\frac{1}{\text{x}^2}=5$ is not a quadratic equation.
View full question & answer→Question 363 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:
$3\text{x}^2-2\sqrt6\text{x}+2=0$
AnswerThe given equation is:
$3\text{x}^2-2\sqrt6\text{x}+2=0$
Comparing it with $ax^2+ bx + c = 0,$ we get
$\text{a}=3,\ \text{b}=-2\sqrt6$ and $\text{c}=2$
$\therefore$ Discriminant, $\text{D}=\text{b}^2-4\text{ac}$
$=\big(-2\sqrt6\big)^2-4\times3\times2$
$=24-24=0$
So, the given equation has real roots.
Now, $\sqrt{\text{D}}=0$
$\therefore\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-\big(2\sqrt6\big)+0}{2\times3}$
$=\frac{2\sqrt6}{6}$
$=\frac{\sqrt6}{3}$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-\big(2\sqrt6\big)-0}{2\times3}$
$=\frac{2\sqrt6}{6}$
$=\frac{\sqrt6}{3}$
Hence, $\frac{\sqrt6}{3}$ is the repeated roots of the given equation.
View full question & answer→Question 373 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:
$\frac{1}{\text{x}}-\frac{1}{\text{x}-2}=3,\ \text{x}\neq0,\ 2$
AnswerThe given equation is:
$\frac{1}{\text{x}}-\frac{1}{\text{x}-2}=3,\ \text{x}\neq0,\ 2$
$\Rightarrow\frac{\text{x}-2-\text{x}}{\text{x}(\text{x}-2)}=3$
$\Rightarrow\frac{-2}{\text{x}^2-2\text{x}}=3$
$\Rightarrow-2=3\text{x}^2-6\text{x}$
$\Rightarrow3\text{x}^2-6\text{x}+2=0$
This equation is of the form $ax^2+ bx + c = 0,$ where
$a = 3, b = -6$ and $c = 2.$
$\therefore$ Discriminant, $D = b^2- 4ac$
$= (-6)^2- 4 × 3 × 2$
$= 36 - 24$
$= 12 > 0$
so, the given equation has real roots.
Now,$\sqrt{\text{D}}=\sqrt{12}=2\sqrt3$
$\therefore\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-(-6)+2\sqrt3}{2\times3}$
$=\frac{6+2\sqrt3}{6}$
$=\frac{3+\sqrt3}{3}$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-(-6)-2\sqrt3}{2\times3}$
$=\frac{6-2\sqrt3}{6}$
$=\frac{3-\sqrt3}{3}$
Hence, $\frac{3+\sqrt3}{3}$ and $\frac{3-\sqrt3}{3}$ are the roots of the given equation.
View full question & answer→Question 383 Marks
Solve the following quadratic equation:
$ x^2-2 a x-\left(4 b^2-a^2\right)=0 $
Answer$ x^2-2 a x-\left(4 b^2-a^2\right)=0 $
$ \Rightarrow x^2-2 a x+\left(a^2-4 b^2\right)=0 $
$ \Rightarrow x^2-2 a x+(a-2 b)(a+2 b)=0 $
$ \Rightarrow x^2-(a-2 b) x-(a+2 b) x+(a-2 b)(a+2 b)=0 $
$ \Rightarrow x[x+(a-2 b)]-(a+2 b)[x-(a-2 b)]=0 $
$ \Rightarrow[x-(a-2 b)][x-(a+2 b)]=0 $
$ \Rightarrow x-(a-2 b)=0 \text { or } x-(a+2 b)=0 $
$ \Rightarrow x=a-2 b \text { or } x=a+2 b$
View full question & answer→Question 393 Marks
Solve the following quadratic equation:
$ 4 x^2-9 x=100 $
Answer$ 4 x^2-9 x=100 $
$ \Rightarrow 4 x^2-9 x-100=0 $
$ \Rightarrow 4 x^2-25 x+16 x-100=0 $
$ \Rightarrow x(4 x-25)+4(4 x-25)=0 $
$ \Rightarrow(4 x-25)(x+4)=0 $
$ \Rightarrow 4 x-25=0 \text { or } x+4=0$
$\Rightarrow\text{x}=\frac{25}{4}$ or $x = -4$
Hence, $\frac{25}{4}$ and $-4$ are the roots of the equation $ 4 x^2-9 x=100 $
View full question & answer→Question 403 Marks
Find the values k for which of roots of $9 x^2+8 k x+16=0$ are real and equal
AnswerGiven:
$9 x^2+8 k x+16=0$
Here,
$a=9, b=8 k \text { and } c=16$
It is given that the roots of the equation are real and equal; therefore, we have:
$ D=0 $
$ \Rightarrow\left(b^2-4 a c\right)=0 $
$ \Rightarrow(8 k)^2-4 \times 9 \times 16=0 $
$ \Rightarrow 64 k^2-576=0 $
$ \Rightarrow 64 k^2=576 $
$ \Rightarrow k^2=9 $
$ \Rightarrow k= \pm 3 $
$ \therefore k=3 \text { or } k=-3$
View full question & answer→Question 413 Marks
Solve the following quadratic equation:
$ 9 x^2+6 x+1=0 $
Answer$ 9 x^2+6 x+1=0 $
$ \Rightarrow 9 x^2+3 x+3 x+1=0 $
$ \Rightarrow 3 x(3 x+1)+1(3 x+1)=0 $
$ \Rightarrow(3 x+1)(3 x+1)=0 $
$ \Rightarrow(3 x+1)^2=0 $
$ \Rightarrow 3 x+1=0$
$\Rightarrow\text{x}=\frac{-1}{3}$ (repeated root)
View full question & answer→Question 423 Marks
Solve the following quadratic equation:
$ x^2+6 x-\left(a^2+2 a-8\right)=0 $
Answer$ x^2+6 x-\left(a^2+2 a-8\right)=0 $
$ \Rightarrow x^2+6 x-\left(a^2+4 a-2 a-8\right)=0 $
$ \Rightarrow x^2+6 x-[a(a+4)-2(a+4)]=0 $
$ \Rightarrow x^2+6 x-(a+4)(a-2)=0 $
$ \Rightarrow x^2+(a+4) x-(a-2) x-(a+4)(a-2)=0 $
$ \Rightarrow x[x+(a+4)]-(a-2)[x+(a+4)]=0 $
$ \Rightarrow[x+(a+4)][x-(a-2)]=0 $
$ \Rightarrow x+(a+4)=0 \text { or } x-(a-2)=0 $
$ \Rightarrow x=-(a+4) \text { or } x=(a-2)$
View full question & answer→Question 433 Marks
The following are quadratic equations in $x?$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=2\Big(\text{x}+\frac{1}{\text{x}}\Big)+3$
Answer$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=2\Big(\text{x}+\frac{1}{\text{x}}\Big)+3$
$\Rightarrow\text{x}^2+2+\frac{1}{\text{x}^2}=\text{2x}+\frac{2}{\text{x}}+3$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}-\text{2x}-\frac{2}{\text{x}}-1=0$
$\Rightarrow\text{x}^4+1-\text{2x}^3-\text{2x}-\text{x}^2=0$
$\Rightarrow\text{x}^4-\text{2x}^3-\text{x}^2-\text{2x}+1=0$
Clearly, $x^4-2 x^3-x^2-2 x+1$ is a polynomial of degree $4$
This is not of the form $a x^2+b x+c=0$
Hence, the given equation is not a quadratic equation.
View full question & answer→Question 443 Marks
Solve the following quadratic equation:
$\text{x}^2-\big(1+\sqrt2\big)\text{x}+\sqrt2=0$
Answer$\text{x}^2-\big(1+\sqrt2\big)\text{x}+\sqrt2=0$
$\Rightarrow\text{x}^2-1.\text{x}-\sqrt2\text{x}+\sqrt2=0$
$\Rightarrow\text{x}(\text{x}-1)-\sqrt2(\text{x}-1)=0$
$\Rightarrow(\text{x}-1)\big(\text{x}-\sqrt2\big)=0$
$\Rightarrow(\text{x}-1)=0$ or $\text{x}-\sqrt2=0$
$\Rightarrow\text{x}=1$ or $\text{x}=\sqrt2$
Hence, $1$ and $\sqrt2$ are the roots of the given equation.
View full question & answer→Question 453 Marks
Solve the following quadratic equation:
$\text{x}^2+2\sqrt{2}\text{x}-6=0$
Answer$\text{x}^2+2\sqrt{2}\text{x}-6=0$
$\Rightarrow\text{x}^2+3\sqrt{2}\text{x}-\sqrt{2}\text{x}-6=0$
$\Rightarrow\text{x}\big(\text{x}+3\sqrt{2}\big)-\sqrt{2}\big(\text{x}+3\sqrt{2}\big)$
$\Rightarrow\big(\text{x}+3\sqrt{2}\big)\big(\text{x}-\sqrt2\big)=0$
$\Rightarrow\text{x}=-3\sqrt2$ or $\text{x}=\sqrt2$
View full question & answer→Question 463 Marks
Solve the following quadratic equation:
$\sqrt3\text{x}^2-2\sqrt2\text{x}-2\sqrt3=0$
Answer$\sqrt3\text{x}^2-2\sqrt2\text{x}-2\sqrt3=0$
$\Rightarrow\sqrt{3}\text{x}^2-3\sqrt2\text{x}-\sqrt{2}\text{x}-2\sqrt{3}=0$
$\Rightarrow\sqrt3\text{x}\big(\text{x}-\sqrt6\big)-\sqrt2\big(\text{x}-\sqrt6\big)=0$
$\Rightarrow\big(\text{x}-\sqrt6\big)\big(\sqrt3\text{x}+\sqrt2\big)=0$
$\Rightarrow\text{x}-\sqrt6=0$ or $\sqrt{3}\text{x}+\sqrt2=0$
$\Rightarrow\text{x}=\sqrt{6}$ or $\text{x}=\frac{-\sqrt{2}}{\sqrt{3}}$
View full question & answer→Question 473 Marks
Find the values of $k$ for which the quadratic equation $(3 k+1) x^2+2(k+1) x+1=0$ has real and equal roots.
AnswerThe given equation is $(3 k+1) x^2+2(k+1) x+1=0$.
This is of the form $a x^2+b x+c=0$, where $a=3 k+1, b=2(k+1)$ and $c=1$.
$ \therefore D=b^2-4 a c $
$ D=[2(k+1)]^2-4 \times(3 k+1) \times 1 $
$ D=\left(k^2+2 k+1\right)-4(3 k+1) $
$ D=4 k^2+8 k+4-12 k-4 $
$ D=4 k^2-4 k$
The given equation will have real and equal roots if $D=0$.
$ \therefore 4 k^2-4 k=0 $
$ \Rightarrow 4 k(k-1)=0 $
$ \Rightarrow k=0 \text { or } k-1=0$
Hence, $0$ and $1$ are the required values of $k$.
View full question & answer→Question 483 Marks
Solve the following quadratic equation:
$\frac{2}{\text{x}^2}-\frac{5}{\text{x}}+2=0$
Answer$\frac{2}{\text{x}^2}-\frac{5}{\text{x}}+2=0$
$\text { Multiplying by } x^2$
$ 2-5 x+2 x^2=0 \text { or } 2 x^2-5 x+2=0 $
$ \Rightarrow 2 x^2-4 x-x+2=0 $
$ \Rightarrow 2 x(x-2)-1(x-2)=0 $
$ \Rightarrow(x-2)(2 x-1)=0 $
$ \therefore x-2=0 \text { or } 2 x-1=0$
$⇒ x = 2, \text{x}=\frac{1}{2}$
Hence, $2$ and $\frac{1}{2}$ are the roots of the given equation.
View full question & answer→Question 493 Marks
If $3$ is a root of the quadratic equation $x^2-x+k=0$, find the value of $p$ so that the roots of the equation $x^2+ k(2x + k + 2) + p = 0$ are equal.
AnswerIt is given that $3$ is a root of the quadratic equation $x^2-x+k=0$.
$ \therefore(3)^2-3+k=0 $
$ \Rightarrow k+6=0 $
$ \Rightarrow k=-6$
The roots of the equation $x^2+2 k x+\left(k^2+2 k+p\right)=0$ are equal.
$\therefore D=0 $
$ \Rightarrow(2 k)^2-4 \times 1 \times\left(k^2+2 k+p\right)=0 $
$ \Rightarrow 4 k 2-4 k 2-8 k-4 p=0 $
$ \Rightarrow-8 k-4 p=0 $
$ \Rightarrow p=\frac{8 k}{-4}=-2 k $
$ \Rightarrow p=-2 \times(-6)=12$
Hence, the value of $p$ is $12 .$
View full question & answer→Question 503 Marks
Solve the following equations by using the method of completing the square:
$\frac{2}{\text{x}^2}-\frac{5}{\text{x}}+2=0$
Answer$\frac{2}{\text{x}^2}-\frac{5}{\text{x}}+2=0$
$\Rightarrow\frac{2-\text{5x}+\text{2x}^2}{\text{x}^2}=0$
$ \Rightarrow 2 x^2-5 x+2=0 $
$ \left.\Rightarrow 4 x^2-10 x+4=0 \text { (Multiplying both sides by } 2\right) $
$ \Rightarrow 4 x^2-10 x=-4$
$\Rightarrow(\text{2x})^2-2\times\text{2x}\times\frac{5}{2}+\Big(\frac{5}{2}\Big)^2\\=-4+\Big(\frac{5}{2}\Big)^2$ $\Big[$Adding $\Big(\frac{5}2{}\Big)^2$ on both sides$\Big]$
$\Rightarrow\Big(\text{2x}-\frac{5}{2}\Big)^2$
$=-4+\frac{25}{4}$
$=\frac{9}{4}=\Big(\frac{3}{2}\Big)^2$
$\Rightarrow\text{2x}-\frac{5}{2}=\pm\frac{3}{2}$ (Taking square root on both sides)
$\Rightarrow\text{2x}-\frac{5}{2}=\frac{3}{2}$ or $\text{2x}-\frac{5}{2}=-\frac{3}{2}$
$\Rightarrow\text{2x}=\frac{3}{2}+\frac{5}{2}=\frac{8}{2}=4$ or $\text{2x}=-\frac{3}{2}+\frac{5}{2}=\frac{2}{2}=1$
$⇒ x = 2$ or $\text{x}=\frac{1}{2}$
Hence, $2$ and $\frac{1}{2}$ are the roots of the given equation.
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