Question 12 Marks
Solve the following quadratic equation:
$\sqrt7\text{x}^2-\text{6x}-13\sqrt7=0$
Answer$\sqrt7\text{x}^2-\text{6x}-13\sqrt7=0$
$\Rightarrow\sqrt7\text{x}^2-13\text{x}+7\text{x}-13\sqrt7=0$
$\Rightarrow\text{x}\big(\sqrt7\text{x}-13\big)+\sqrt7\big(\sqrt7\text{x}-13\big)=0$
$\Rightarrow\big(\text{x}+\sqrt7\big)\big(\sqrt7\text{x}-13\big)=0$
$\Rightarrow\text{x}+\sqrt7=0$ or $\sqrt7\text{x}-13=0$
$\Rightarrow\text{x}=-\sqrt7$ or $\text{x}=\frac{13\sqrt7}{7}$
Hence, $-\sqrt7$ and $\frac{13\sqrt7}{7}$ are the roots of the given equation.
View full question & answer→Question 22 Marks
Find the value of $\alpha$ for which the equation $(\alpha+12)\text{x}^2+2(\alpha-12)\text{x}+2=0$ has equal roots.
AnswerGiven:
$(\alpha+12)\text{x}^2+2(\alpha-12)\text{x}+2=0$
Here,
$\text{a}=(\alpha-12),\ \text{b}=2(\alpha-12)$ and $\text{c}=2$
It is given that the roots of the given equation are equal; therefore, we have:
$\text{D}=0$
$\Rightarrow(\text{b}^2-\text{4ac})=0$
$\Rightarrow\big\{2(\alpha-12)\big\}^2-4\times(\alpha-12)\times2=0$
$\Rightarrow4(\alpha^2-24\alpha+144)-8(\alpha-12)=0$
$\Rightarrow4\alpha^2-96\alpha+576-8\alpha+96=0$
$\Rightarrow4\alpha^2-104\alpha+672=0$
$\Rightarrow\alpha^2-26\alpha+168=0$
$\Rightarrow\alpha^2-14\alpha-12\alpha+168=0$
$\Rightarrow\alpha(\alpha-14)-12(\alpha-14)=0$
$\Rightarrow(\alpha-14)(\alpha-12)=0$
$\therefore\ \alpha=14$ or $\alpha=12$
If the value of $\alpha$ is $12$, the given equation becomes non-quadratic.Therefore, the value of $\alpha$ will be $14$ for the equation to have equal roots.
View full question & answer→Question 32 Marks
Find the discriminant of the following equation:
$(x - 1)(2x - 1) = 0$
Answer$(x - 1)(2x - 1) = 0$
$\Rightarrow 2 x^2-3 x+1=0$
$\text { Comparing it with } a x^2+b x+c=0 \text {, we get }$
$ a=2, $
$ b=-3, $
$ c=1 $
$ \therefore \text { Discriminant, } D=b^2-4 a c $
$ =(-3)^2-4 \times 2 \times 1$
$= 9 - 8$
$= 1$
View full question & answer→Question 42 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:
$25 x^2+30 x+7=0$
AnswerGiven,
$25 x^2+30 x+7=0$
On comparing it with $a x^2+b x+c=0$, we get:
$a=25, b=30 \text { and } c=7$
Discriminant $D$ is given by:
$ D=\left(b^2-4 a c\right) $
$=30^2-4 \times 25 \times 7$
$= 900 - 700$
$= 200$
$= 200 > 0$
Hence, the roots of the equation are real.
Roots $\alpha$ and $\beta$ are given by:
$\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-30+\sqrt{200}}{2\times25}$
$=\frac{-30+10\sqrt2}{50}$
$=\frac{10\big(-3+\sqrt2\big)}{50}$
$=\frac{\big(-3+\sqrt2\big)}{5}$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-30-\sqrt{200}}{2\times25}$
$=\frac{-30-10\sqrt2}{50}$
$=\frac{10\big(-3-\sqrt2\big)}{50}$
$=\frac{\big(-3-\sqrt5\big)}{5}$
Thus, the roots of the equation are $\frac{-3+\sqrt2}{5}$ and $\frac{-3-\sqrt2}{5}$
View full question & answer→Question 52 Marks
The following are quadratic equations in x?
$(2x + 3)(3x + 2) = 6(x - 1)(x - 2)$
Answer$(2x + 3)(3x + 2) = 6(x - 1)(x - 2)$
$ \Rightarrow 6 x^2+4 x+9 x+6=6\left(x^2-2 x-x+2\right) $
$ \Rightarrow 6 x^2+13 x+6=6 x^2-18 x+2$
$⇒ 13x + 18x + 6 - 2 = 0$
$⇒ 31x + 4 = 0$
This is not of the form $ax^2+ bx + c = 0$
Hence, the given equation is not a quadratic equation.
View full question & answer→Question 62 Marks
Find the values of p for which the quadratic equation $(2 p+1) x^2-(7 p+2) x+(7 p-3)=0$ has real and equal roots.
AnswerThe given equation is $(2 p+1) x^2-(7 p+2) x+(7 p-3)=0$
This is of the form $a x^2+b x+c=0$, where $a=2 p+1, b=-(7 p+2)$ and $c=7 p-3$.
$ \therefore D=b^2-4 a c $
$ D=\left[-(7 p+2)^2-4 \times(2 p+1) \times(7 p-3)\right] $
$ D=\left(49 p^2+28 p+4\right)-4\left(14 p^2+p-3\right) $
$ D=49 p^2+28 p+4-56 p^2-4 p+12 $
$ D=-7 p^2+24 p+16$
The given equation will have real and equal roots if $D=0$.
$ \therefore-7 p^2+24 p+16=0 $
$ \Rightarrow 7 p^2-24 p-16=0 $
$ \Rightarrow 7 p^2-28 p+4 p-16=0 $
$ \Rightarrow 7 p(p-4)+4(p-4)=0 $
$ \Rightarrow(p-4)(7 p+4)=0 $
$ \Rightarrow p-4=0 \text { or } 7 p+4=0$
$\Rightarrow p = 4$ or $\text{p}=-\frac{4}{7}$
Hence, $4$ and $-\frac{4}{7}$ are the required values of $p$.
View full question & answer→Question 72 Marks
Find the nature of the roots of the following quadratic equations:
$\text{12x}^2-4\sqrt{15}\text{x}+5=0$
AnswerThe given equation is $\text{12x}^2-4\sqrt{15}\text{x}+5=0$
This is of the form $ax^2+ bx + c = 0,$ where $\text{a}=12,\ \text{b}=-4\sqrt{15}$ and $c = 5$
$\therefore$ Discriminant, $D = b^2- 4ac$
$\text{D}=\big(-4\sqrt{15}\big)^2-4\times12\times5$
$\text{D}=240-240=0$
Hence, the given equation has real and equal roots.
View full question & answer→Question 82 Marks
Solve:
$ 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=0$
$⇒ 2x(x + a) - a(x + a) = 0$
$⇒ (x + a)(2x - a) = 0$
$⇒ x + a = 0 or 2x - a = 0$
$⇒ x = -a$ or $\text{x}=\frac{\text{a}}2{}$
View full question & answer→Question 92 Marks
Show that the roots of the equation $x^2+p x-q^2=0$ are real for all real value of $p$ and $q$.
AnswerGiven:
$x^2+p x-q^2=0$
Here,
$a=1, b=p \text { and } c=-q^2$
Discriminant $D$ is given by:
$ D=\left(b^2-4 a c\right) $
$ D=p^2-4 \times 1 \times\left(-q^2\right) $
$ D=\left(p^2+4 q^2\right)>0$
$D > 0$ for all real values of $p$ and $q$.
Thus, the roots of the equation are real.
View full question & answer→Question 102 Marks
Find the nature of the roots of the following quadratic equations:
$\text{3x}^2-2\sqrt6\text{x}+2=0$
AnswerThe given equation is $\text{3x}^2-2\sqrt6\text{x}+2=0$
This is of the form $ax^2+ bx + c = 0,$ where $\text{a}=3,\ \text{b}=-2\sqrt6$ and $c = 2$
$\therefore$ Discriminant, $D = b^2- 4ac$
$\text{D}=\big(2\sqrt6\big)^2-4\times3\times2$
$\text{D}=24-24=0$
Hence, the given equation has real and equal roots.
View full question & answer→Question 112 Marks
Find the nature of the roots of the following quadratic equations:
$2x^2- 8x + 5 = 0$
AnswerThe given equation is $2 x^2-8 x+5=0$
This is of the form $\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=0$, where $\mathrm{a}=2, \mathrm{~b}=-8$ and $\mathrm{c}=5$
$\therefore$ Discriminant, $D=b^2-4 a c$
$D = (-8)^2- 4 × 2 × 5$
$D = 64 - 40$
$D = 24 > 0$
Hence, the given equation has real and unequal roots.
View full question & answer→Question 122 Marks
Find the discriminant of the following equation:
$\sqrt3\text{x}^2+2\sqrt2\text{x}-2\sqrt3=0$
Answer$\sqrt3\text{x}^2+2\sqrt2\text{x}-2\sqrt3=0$
Here,
$\text{a}=\sqrt3,$
$\text{b}=2\sqrt2,$
$\text{c}=-2\sqrt3$
Discriminant $D$ is given by:
$\text{D}=\text{b}^2-4\text{ac}$
$=\big(2\sqrt2\big)^2-4\times\sqrt3\times\big(-2\sqrt3\big)$
$=(4\times2)+(8\times3)$
$=8+24$
$=32$
View full question & answer→Question 132 Marks
Find the nature of the roots of the following quadratic equations:
$5 x^2-4 x+1=0$
AnswerThe given equation is $5 x^2-4 x+1=0$
This is of the form $a x^2+b x+c=0$, where $a=5, b=-4$ and $c=1$
$\therefore$ Discriminant, $D=b^2-4 a c$
$D=(-4)^2-4 \times 5 \times 1$
$D = 16 - 20$
$D = -4 > 0$
Hence, the given equation has no real roots.
View full question & answer→Question 142 Marks
Find the discriminant of the following equation:
$3x^2- 2x + 8 = 0$
Answer$3x^2- 2x + 8 = 0$
Here,
$a = 3,$
$b = -2,$
$c = 8$
Discriminant $D$ is given by:
$D = b^2- 4ac$
$D = (-2)^2- 4 × 3 × 8$
$D = 4 - 96$
$D = -92$
View full question & answer→Question 152 Marks
Solve the following quadratic equation:
$ x^2+12 x+35=0 $
Answer$ x^2+12 x+35=0 $
$ \Rightarrow x^2+7 x+5 x+35=0 $
$⇒ x(x + 7) + 5(x + 7) = 0$
$⇒ (x + 7)(x + 5) = 0$
$⇒ x + 7 = 0 or x + 5 = 0$
$⇒ x = -7 or x = -5$
View full question & answer→Question 162 Marks
Solve the following quadratic equation:
$ 3 x^2-243=0 $
Answer$ 3 x^2-243=0 $
$ \Rightarrow 3\left(x^2-81\right)=0$
$\Rightarrow x^2=81 $
$\Rightarrow\text{x}=\pm\sqrt{81}=\pm9$
$⇒ x = 9, -9$
Hence 9 and -9 are the roots of the equation $3x^2- 243 = 0$
View full question & answer→Question 172 Marks
Solve the following quadratic equation:
$4 x^2+5 x=0$
Answer$4 x^2+5 x=0$
$⇒ x(4x + 5) = 0$
$⇒ x = 0$ or $4x + 5 = 0$
$⇒ x = 0$ or $\text{x}=-\frac{5}{4}$
View full question & answer→Question 182 Marks
Find the values of $p$ for which the quadratic equation $(p + 1)x^2- 6(p + 1)x + 3(p + 9) = 0, p ≠ -1$ has equal roots. Hence, find the roots of the equation.
AnswerThe given equation is $(p + 1)x^2- 6(p + 1)x + 3(p + 9) = 0$
This is of the form $ax^2+ bx + c = 0,$ where $a = p + 1, b = -6(p + 1)$ and $c = 3(p + 9).$
$ \therefore D=b^2-4 a c $
$ \left.D=[-6(p+1)]^2-4 \times(p+1) \times 3(p+9)\right] $
$ D=12(p+1)[3(p+1)-(p+9)] $
$ D=12(p+1)(2 p-6)$
The given equation will have real and equal roots if $D = 0$.
$\therefore 12(p + 1)(2p - 6) = 0$
$\Rightarrow p + 1 = 0$ or $2p - 6 = 0$
$\Rightarrow p = -1$ or $p = 3$
But, $p ≠ -1$ (Given)
Thus, the value of $p$ is $3$
Putting $p = 3$, the given equation becomes $4x^2- 24x + 36 = 0.$
$ 4 x^2-24 x+36=0 $
$ \Rightarrow 4\left(x^2-6 x+9\right)=0 $
$ \Rightarrow(x-3)^2=0 $
$ \Rightarrow x-3=0 $
$ \Rightarrow x=3$
Hence, $3$ is the repeated root of this equation.
View full question & answer→Question 192 Marks
The following are quadratic equations in $x?$
$\text{x}^2-3\text{x}-\sqrt{\text{x}}+4=0$
Answer$\text{x}^2-3\text{x}-\sqrt{\text{x}}+4$ is not a quadratic polynomial since it contains $\sqrt{\text{x}}$ or $\text{x}^{\frac{1}{2}}$ in which power $\frac{1}{2}$ of x is not an integer.
$\therefore\ \text{x}^2-3\text{x}-\sqrt{\text{x}}+4=0$ is a quadratic equation.
View full question & answer→Question 202 Marks
Solve: $ 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)[2 x+(b-a)]=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+(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 212 Marks
If $a$ and $b$ are distinct real numbers, show that the quadratic equation $2\left(a^2+b^2\right) x^2+2(a+b) x+1=0$ has no real roots.
AnswerThe given equation is $2\left(a^2+b^2\right) x^2+2(a+b) x+1=0$
$ \therefore D=[2(a+b)]^2-4 \times 2\left(a^2+b^2\right) \times 1 $
$ =4\left(a^2+2 a b+b^2\right)-8\left(a^2+b^2\right) $
$ =4 a^2+8 a b+4 b^2-8 a^2-8 b^2 $
$ =-4 a^2+8 a b-4 b^2 $
$ =-4\left(a^2-2 a b+b^2\right) $
$ =-4(a-b)^2<0$
Hence, the given equation has no real roots.
View full question & answer→Question 222 Marks
The following are the roots of $3 x^2+2 x-1=0?$
$\frac{1}3{}$
AnswerThe given equation is $3 x^2+2 x-1=0$
On substituting $\text{x}=\frac{1}{3}$ in the equation, we get
$\text{LHS}=3\times\Big(\frac{1}{3}\Big)^2+2\times\Big(\frac{1}{3}\Big)-1=0$
$=\Big(\frac{1}{3}+\frac{2}{3}-1\Big)=0=\text{RHS}$
$\therefore\text{x}=\frac{1}{3}$ is a solution of $3 x^2+2 x-1=0$
View full question & answer→Question 232 Marks
Solve the following quadratic equation:
$\frac{16}{\text{x}}-1=\frac{15}{\text{x}+1},$ $\text{x}\neq0,\ 0,\ 1$
Answer$\frac{16}{\text{x}}-1=\frac{15}{\text{x}+1}$
$\Rightarrow\frac{16-\text{x}}{\text{x}}=\frac{15}{\text{x}+1}$
$ \Rightarrow(16-x)(x+1)=15 x $
$ \Rightarrow 16 x+16-x^2-x=15 x $
$ \Rightarrow 15 x+16-x^2=15 x $
$ \Rightarrow x^2-16=0 $
$ \Rightarrow(x-4)(x+4)=0 $
$ \Rightarrow x-4=0 \text { or } x+4=0 $
$ \Rightarrow x=4 \text { or } x=-4$
View full question & answer→Question 242 Marks
Solve the following quadratic equation:
$\text{x}^2-\big(\sqrt3+1\big)\text{x}+\sqrt3=0$
Answer$\text{x}^2-\big(\sqrt3+1\big)\text{x}+\sqrt3=0$
$\Rightarrow\text{x}^2-\sqrt3\text{x}-\text{x}+\sqrt3=0$
$\Rightarrow\text{x}\big(\text{x}-\sqrt3\big)-1\big(\text{x}-\sqrt3\big)=0$
$\Rightarrow\big(\text{x}-\sqrt3\big)\big(\text{x}-1\big)=0$
$\Rightarrow\text{x}-\sqrt3=0$ or $\text{x}-1=0$
$\Rightarrow\text{x}=\sqrt{3}$ or $\text{x}=1$
View full question & answer→Question 252 Marks
Solve the following quadratic equation:
$ x^2+6 x+5=0$
Answer$ x^2+6 x+5=0$
$ \Rightarrow x^2+5 x+x+5=0 $
$⇒ x(x + 5) + 1(x + 5) = 0$
$⇒ (x + 5)(x + 1) = 0$
$⇒ x + 5 = 0 or x + 1 = 0$
$⇒ x = -5 or x = -1$
View full question & answer→Question 262 Marks
Show that $\text{x}=-\frac{\text{bc}}{\text{ad}}$ is a solution of the quadratic equation $\text{ad}^2\Big(\frac{\text{ax}}{\text{b}}+\frac{\text{2c}}{\text{d}}\Big)\text{x}+\text{bc}^2=0$
Answer$\text{ad}^2\Big(\frac{\text{ax}}{\text{b}}+\frac{\text{2c}}{\text{d}}\Big)\text{x}+\text{bc}^2=0$
By multiplying $ad^2x$ by $\frac{\text{ax}}{\text{b}}$ we get
$\text{ad}^2\text{x}\Big(\frac{\text{ax}}{\text{b}}\Big)$
$=\frac{\text{a}^2\text{d}^2\text{x}^2}{ \text{b}}$
By multiplying $ad^2x$ by $\frac{\text{2c}}{\text{d}}$ we get
$\text{ad}^2\text{x}\Big(\frac{\text{2c}}{\text{d}}\Big)$
$=\frac{\text{2ca}\text{d}^2\text{x}}{\text{d}}$
By following the equation
$\frac{\text{a}^2\text{d}^2\text{x}^2}{\text{b}}+\frac{\text{2ca}\text{d}^2\text{x}}{\text{d}}+\text{bc}^2=0$
$\frac{\text{a}^2\text{d}^2\text{x}^2}{\text{b}}+\text{2cad}^{2-1}\text{x}+\text{bc}^2=0$
$\frac{\text{a}^2\text{d}^2\text{x}^2}{\text{b}}+\text{2cad}\text{x}+\text{bc}^2=0$
$a^2 d^2 x+2 a b c d x+b^2 c^2=0$ (by multiplying the denominator $'b'$ to $2cadx$ & $b c^2$)
$\Rightarrow (adx + bc) = 0$ (Squaring)
$\Rightarrow adx + bc = 0$
$\Rightarrow adx = -bc$
$\Rightarrow\text{x}=\frac{-\text{bc}}{\text{ad}}$
View full question & answer→Question 272 Marks
Find the discriminant of the following equation:
$2x^2- 7x + 6 = 0$
Answer$2 x^2-7 x+6=0$
Here,
$a = 2,$
$b = -7,$
$c = 6$
Discriminant $D$ is given by:
$D=b^2-4 a c$
$D=(-7)^2-4 \times 2 \times 6$
$D = 49 - 48$
$D = 1$
View full question & answer→Question 282 Marks
Solve the following quadratic equation:
$3\sqrt7\text{x}^2+\text{4x}-\sqrt7=0$
Answer$3\sqrt7\text{x}^2+\text{4x}-\sqrt7=0$
$\Rightarrow3\sqrt7\text{x}^2+7\text{x}-3\text{x}-\sqrt7=0$
$\Rightarrow\sqrt7\text{x}\big(3\text{x}+\sqrt7\big)-1\big(3\text{x}+\sqrt7\big)=0$
$\Rightarrow\big(\text{3x}+\sqrt{7}\big)\big(\sqrt7\text{x}-1\big)=0$
$\Rightarrow\text{3x}+\sqrt7=0$ or $\sqrt7\text{x}-1=0$
$\Rightarrow\text{3x}=-\sqrt7$ or $\text{x}=\frac{1}{\sqrt7}$
$\Rightarrow\text{x}=\frac{-\sqrt7}{3}$ or $\text{x}=\frac{1\times\sqrt7}{\sqrt7\times\sqrt7}=\frac{\sqrt7}{7}$
Hence, $\frac{-\sqrt7}{3}$ and $\frac{\sqrt7}{7}$ are the roots of the given equation.
View full question & answer→Question 292 Marks
$ 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 302 Marks
Find the value of k for which the equation $x^2+k(2 x+k-1)+2=0$ has real and equal roots.
AnswerThe given equation is $x^2+k(2 x+k-1)+2=0$.
$\Rightarrow x^2+2 k x+k(k-1)+2=0$
So, $a=1, b=2 k, c=k(k-1)+2$
We know $D=b^2-4 a c$
$ \Rightarrow D=(2 k)^2-4 \times 1 \times[k(k-1)+2] $
$ \Rightarrow D=4 k^2-4\left[k^2-k+2\right]$
$ \Rightarrow D=4 k^2-4 k^2+4 k-8$
$⇒ D = 4k - 8 = 4(k - 2)$
For equal roots,$D = 0$
Thus, $4(k - 2) = 0$
So, $k = 2.$
View full question & answer→Question 312 Marks
Solve: $4\sqrt3\text{x}^2+5\text{x}-2\sqrt3=0$
Answer$4\sqrt3\text{x}^2+5\text{x}-2\sqrt3=0$
$\Rightarrow4\sqrt3\text{x}^2+\text{8x}-\text{3x}-2\sqrt3=0$
$\Rightarrow\text{4x}\big(\sqrt3\text{x}+2\big)-\sqrt3\big(\sqrt3\text{x}+2\big)=0$
$\Rightarrow\big(\sqrt3\text{x}+2\big)\big(4\text{x}-\sqrt3\big)=0$
$\Rightarrow\sqrt3\text{x}+2=0$ or $4\text{x}-\sqrt3=0$
$\Rightarrow\text{x}=-\frac{2}{\sqrt3}=-\frac{2\sqrt3}{3}$ or $\text{x}=\frac{\sqrt3}{4}$
View full question & answer→Question 322 Marks
For what values of $p$ are the roots of the equation $4x^2+ px + 3 = 0$ real and equal?
AnswerThe given equation is $4x^2+ px + 3 = 0.$
This is of the form $ax^2+ bx + c = 0$, where $a = 4, b = p$ and $c = 3.$
$\therefore D = b^2- 4ac$
$D = p^2- 4 × 4 × 3$
$D = p^2- 48$
The given equation will have real and equal roots if $D = 0$.
$\therefore p^2- 48 = 0$
$\Rightarrow p^2= 48$
$\Rightarrow\text{p}=\pm48=\pm4\sqrt3$
Hence, $4\sqrt3$ and $-4\sqrt3$ are the required values of $p$.
View full question & answer→Question 332 Marks
Find the nature of the roots of the following quadratic equations:
$x^2-x+2=0$
AnswerThe given equation is $x^2-x+2=0$
This is of the form $a x^2+b x+c=0$, where $a=1, b=-1$ and $c=2$
$\therefore$ Discriminant, $\mathrm{D}=\mathrm{b}^2-4 \mathrm{ac}$
$D=(-1)^2-4 \times 1 \times 2$
$D=1-8$
$D=-7<0$
Hence, the given equation has no real roots.
View full question & answer→Question 342 Marks
Solve: $ 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 352 Marks
Solve the following quadratic equation:
$ 15 x^2-28=x $
Answer$ 15 x^2-28=x $
$ \Rightarrow 15 x^2-x-28=0 $
$ \Rightarrow 15 x^2-21 x+20 x-28=0 $
$ \Rightarrow 3 x(5 x-7)+4(5 x-7)=0 $
$ \Rightarrow(5 x-7)(3 x+4)=0 $
$ \Rightarrow 5 x-7=0 \text { or } 3 x+4=0$
$\Rightarrow\text{x}=\frac{7}{5}$ or $\text{x}=\frac{-4}{3}$
Hence, $\frac{7}{5}$ and $\frac{-4}{3}$ are the roots of the equation $ 15 x^2-28=x $
View full question & answer→Question 362 Marks
Find the discriminant of the following equation:
$\text{2x}^2-5\sqrt{2\text{x}}+4=0$
Answer$\text{2x}^2-5\sqrt{2\text{x}}+4=0$
Here,
$\text{a} = 2,$
$\text{b}=-5\sqrt2,$
$\text{c}=4$
Discriminant $D$ is given by:
$\text{D} = \text{b}^2 - \text{4ac}$
$\text{D}=\big(-5\sqrt2\big)^2-4\times2\times4$
$\text{D} = 49 - 48$
$\text{D} = 1$
View full question & answer→Question 372 Marks
Solve the following quadratic equation:
$ 6 x^2+x-12=0 $
Answer$ 6 x^2+x-12=0 $
$ \Rightarrow 6 x^2+9 x-8 x-12=0 $
$ \Rightarrow 3 x(2 x+3)-4(2 x+3)=0 $
$ \Rightarrow(2 x+3)(3 x-4)=0 $
$ \Rightarrow 2 x+3=0 \text { or } 3 x-4=0$
$\Rightarrow\text{x}=\frac{-3}{2}$ or $\text{x}=\frac{4}{3}$
Hence, $\frac{-3}{2}$ and $\frac{4}{3}$ are the roots of $ 6 x^2+x-12=0 $
View full question & answer→Question 382 Marks
Solve the following quadratic equation:
$ 2 x^2+x-6=0 $
Answer$ 2 x^2+x-6=0 $
$ \Rightarrow 2 x^2+4 x-3 x-6=0 $
$ \Rightarrow 2 x(x+2)-3(x+2)=0 $
$ \Rightarrow(x+2)(2 x-3)=0 $
$ \Rightarrow x+2=0 \text { or } 2 x-3=0$
$⇒ x = -2$ or $\text{x}=\frac{3}{2}$
View full question & answer→Question 392 Marks
$ x^2-4 a x+4 a^2-b^2=0 $
Answer$ x^2-4 a x+4 a^2-b^2=0 $
$ \Rightarrow x^2-4 a x+\left(4 a^2-b^2\right)=0 $
$ \Rightarrow x^2-4 a x+(2 a+b)(2 a-b)=0 $
$ \Rightarrow x^2-(2 a+b) x-(2 a-b) x+(2 a+b)(2 a-b)=0 $
$ \Rightarrow x[x-(2 a+b)]-(2 a-b)[x+(2 a+b)]=0 $
$ \Rightarrow[x-(2 a+b)][x-(2 a-b)]=0 $
$ \Rightarrow x-(2 a+b)=0 \text { or } x-(2 a-b)=0 $
$ \Rightarrow x=2 a+b \text { or } x=2 a-b$
View full question & answer→Question 402 Marks
Solve: $\sqrt3\text{x}^2-2\sqrt2\text{x}-2\sqrt3=0$
Answer$\sqrt3\text{x}^2-2\sqrt2\text{x}-2\sqrt3=0$
$\Rightarrow\sqrt3\text{x}^2-3\sqrt2\text{x}+\sqrt2\text{x}-2\sqrt3=0$
$\Rightarrow\sqrt3\text{x}\big(\text{x}-\sqrt6\big)+\sqrt2\big(\text{x}-\sqrt6\big)=0$
$\Rightarrow\big(\sqrt3\text{x}+\sqrt2\big)\big(\text{x}-\sqrt6\big)=0$
$\Rightarrow\sqrt3\text{x}+\sqrt2$ or $\text{x}-\sqrt6$
$\Rightarrow\text{x}=\frac{-\sqrt2}{\sqrt3}$ or $\text{x}=\sqrt6$
View full question & answer→Question 412 Marks
The following are quadratic equations in $x?$
$\text{x}-\frac{6}{\text{x}}=3$
Answer$\text{x}-\frac{6}{\text{x}}=3$
$\Rightarrow x^2-6=3 x$
$ \Rightarrow x^2-3 x-6=0$
And $\left(x^2-3 x-6\right)$ Being a polynomial of degree $2$, it is a quadratic polynomial.
Hence $\text{x}-\frac{6}{\text{x}}=3$ is a quadratic equation.
View full question & answer→Question 422 Marks
For what value of $k$ are the roots of the quadratic equation $\text{kx}\big(\text{x}-2\sqrt5\big)+10=0$ real and equal?
AnswerThe given equation is:
$\text{kx}\big(\text{x}-2\sqrt5\big)+10=0$
$\Rightarrow\text{kx}^2-2\sqrt5\text{kx}+10=0$
This is of the form $ax2 + bx + c = 0$, where $\text{a}=\text{k},\ \text{b}=-2\sqrt5\text{k}$ and $c = 10$
$\therefore\text{D}=\text{b}^2-\text{4ac}$
$\text{D}=\big(-2\sqrt5\text{k}\big)^2-4\times\text{k}\times10$
$\text{D}=\text{20k}^2-\text{40k}$
The given equation will have real and equal roots if $D = 0$.
$\therefore 20k^2- 40k = 0$
$\Rightarrow 20k(k - 2) = 0$
$\Rightarrow k = 0$ or $k - 2 = 0$
$\Rightarrow k = 0$ or $k = 2$
But, for $k = 0$, we get $10 = 0$, which is not true.
Hence, $2$ is the required value of $k$.
View full question & answer→Question 432 Marks
Solve the following quadratic equation:
$4 - 11x = 3x^2$
Answer$ 4-11 x=3 x^2 $
$ \Rightarrow 3 x^2+11 x-4=0 $
$ \Rightarrow 3 x^2+12 x-x-4=0 $
$ \Rightarrow 3 x(x+4)-1(x+4)=0 $
$ \Rightarrow(x+4)(3 x-1)=0 $
$ \Rightarrow x+4=0 \text { or } 3 x-1=0$
$\Rightarrow x = -4$ or $\text{x}=\frac{1}{3}$
Hence, $-4$ and $\frac{1}{3}$ are the roots of the equation $4 - 11x = 3x^2$
View full question & answer→Question 442 Marks
Solve the following quadratic equation:
$(2x - 3)(3x + 1) = 0$
Answer$(2x - 3)(3x + 1) = 0$
$\Rightarrow 2x - 3 = 0$ or $3x + 1 = 0$
$\Rightarrow 2x = 3$ or $3x = -1$
$\Rightarrow\text{x}=\frac{3}{2}$ or $\text{x}=-\frac{1}{3}$
View full question & answer→Question 452 Marks
The following are the roots of $3 x^2+2 x-1=0?$
AnswerThe given equation is $3 x^2+2 x-1=0$
On substituting $x=-1$ in the equation, we get
$ \text { LHS }=3 \times(-1)^2+2 \times(-1)-1 $
$ =3-2-1=0=\text { RHS }$
$\therefore x=-1$ is a solution of $3 x^2+2 x-1=0$
View full question & answer→Question 462 Marks
Solve the following quadratic equation:
$\text{x}^2-3\sqrt3-30=0$
Answer$\text{x}^2-3\sqrt3-30=0$
$\Rightarrow\text{x}^2+5\sqrt3\text{x}-2\sqrt3\text{x}-30=0$
$\Rightarrow\text{x}\big(\text{x}+5\sqrt3\big)-2\sqrt3\big(\text{x}+5\sqrt3\big)=0$
$\Rightarrow\big(\text{x}+5\sqrt3\big)\big(\text{x}-2\sqrt3\big)=0$
$\Rightarrow\text{x}+5\sqrt3=0$ or $\text{x}-2\sqrt3=0$
$\Rightarrow\text{x}=-5\sqrt{3}$ or $\text{x}=2\sqrt3$
View full question & answer→Question 472 Marks
Solve the following quadratic equation:
$ 100 x^2-20 x+1=0 $
Answer$ 100 x^2-20 x+1=0 $
$ \Rightarrow 100 x^2-10 x-10 x+1=0 $
$ \Rightarrow 10 x(10 x-1)-1(10 x-1)=0 $
$ \Rightarrow(10 x-1)(10 x-1)=0 $
$ \Rightarrow(10 x-1)^2=0 $
$ \Rightarrow 10 x-1=0$
$\Rightarrow\text{x}=\frac{1}{10}$ (repeated root)
View full question & answer→Question 482 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:
$2\sqrt3\text{x}^2-5\text{x}+\sqrt3=0$
AnswerThe given equation is:
$2\sqrt3\text{x}^2-5\text{x}+\sqrt3=0$
Comparing it with $ax^2+ bx + c = 0,$ we get
$\text{a}=2\sqrt3,\ \text{b}=-5$ and $\text{c}=\sqrt3$
$\therefore$ Discriminant, $\text{D}=\text{b}^2-4\text{ac}$
$=(-5)^2-4\times2\sqrt3\times\sqrt3$
$=25-24$
$=1>0$
So, the given equation has real roots.
Now, $\sqrt{\text{D}}=\sqrt1=1$
$\therefore\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-(-5)+1}{2\times2\sqrt3}$
$=\frac{6}{4\sqrt3}$
$=\frac{\sqrt3}{2}$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-(-5)-1}{2\times2\sqrt3}$
$=\frac{4}{4\sqrt3}$
$=\frac{\sqrt3}{3}$
Hence, $\frac{\sqrt3}{2}$ and $\frac{\sqrt3}{3}$ are the roots of the given equation.
View full question & answer→Question 492 Marks
The following are quadratic equations in $x?$
$ (x+2)^3=x^3-8 $
Answer$ (x+2)^3=x^3-8 $
$ \Rightarrow x^3+8+6 x(x+2)=x^3-8 $
$ \Rightarrow x^3+8+6 x^2+12 x=x^3-8 $
$ \Rightarrow 6 x^2+12 x+16=0$
$\text { This is of form } a x^2+b x+c \text {, where }$
$a = 6, b = 12$ and $c = 16$
Hence, the given equation is a quadratic equation.
View full question & answer→Question 502 Marks
The following are quadratic equations in $x?$
$\text{x}+\frac{2}{\text{x}}=\text{x}^2$
Answer$\text{x}+\frac{2}{\text{x}}=\text{x}^2$
$ \Rightarrow x^2+2=x^3 $
$ \Rightarrow x^3-x^2-2=0$
And $(x^3- x^2- 2)$ Being a polynomial of degree $3$, it is a quadratic polynomial.
Hence $\text{x}+\frac{2}{\text{x}}=\text{x}^2$ is a quadratic equation.
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