Question 1013 Marks
Solve for x.
$\text{x}+\frac{1}{\text{x}}=3,\text{x}\neq0$
Answer$\text{x}+\frac{1}{\text{x}}=3$
$\text{x}^2+1=3\text{x}$
$\text{x}^3-3\text{x}+1=0$
Here, $\text{a}=1,\text{b}=-3,\text{c}=1$
$\therefore\text{D}=\text{b}^2-4\text{ac}$
$=(-3)^2-4\times1\times1$
$=9-4=5$
$\because\text{D}>0$
$\therefore$ Roots are real
$\therefore\text{x} = {-\text{b} \pm \sqrt{\text{b}^2-4\text{ac}} \over 2\text{a}}$
$=\frac{-(-3)\pm\sqrt{5}}{2\times1}=\frac{(3)\pm\sqrt{5}}{2}$
$\therefore\text{x}=\frac{3+\sqrt{5}}{2},\frac{3-\sqrt{5}}{2}$
View full question & answer→Question 1023 Marks
Solve the following quadratic equations by factorization:
$9 x^2-3 x-2=0$
AnswerWe have
$9 x^2-3 x-2=0$
$\Rightarrow 9 x^2-6 x+3 x-2=0$
$\Rightarrow 3 x(3 x-2)+1(3 x-2)=0$
$\Rightarrow(3 x-2)(3 x+1)=0$
$\Rightarrow \text { either } 3 x-2=0 \text { or } 3 x+1=0$
$\Rightarrow 3 x=2 \text { or } 3 x=-1$
$\Rightarrow\text{x}=\frac{2}{3}$ or $\text{x}=-\frac{1}{3}$
Thus, $\text{x}=\frac{2}{3}$ and $\text{x}=-\frac{1}{3}$ are two roots of the equation $9 x^2-3 x-2=0$
View full question & answer→Question 1033 Marks
A train covers a distance of $90\ km$ at a uniform speed. Had the speed been $15\ km/hr$ more, it would have taken $30$ minutes less for the journey. Find the original speed of the train.
AnswerDistance to be covered $= 90km$
Let uniform-original speed $= x km/h$
Increased speed $= (x + 15) km/hr$
According to the condition,
$\frac{90}{\text{x}}-\frac{90}{\text{x}+15}=\frac{1}{2}$ $\big(30\text{ minutes}=\frac{1}{2}\text{hour}\big)$
$\Rightarrow\frac{90\text{x}+1350-90\text{x}}{\text{x}(\text{x}+15)}=\frac{1}{2}$
$\Rightarrow\frac{1350}{\text{x}^2+15\text{x}}=\frac{1}{2}$
$\Rightarrow\text{x}^2+15\text{x}=2700$
$\Rightarrow\text{x}^2+15\text{x}-2700=0$
$\begin{Bmatrix}\because-2700=60\times(-45)\\15=60-45\end{Bmatrix} $
$⇒ x^2+ 60x - 45x - 2700 = 0$
$⇒ x(x + 60) - 45(x + 60) = 0$
$⇒ (x + 60)(x - 45) =0$
Either $x + 60 = 0, $ then $x = -60$ which is not possible being negative
or $x - 45 = 0$, then $x = 45$
Original speed of the train $= 45\ km/hr$
View full question & answer→Question 1043 Marks
Write the set of values of ‘a’ for which the equation $x^2+a x-1=0$, has real roots.
Answer$x^2+a x-1=0$
$\text { Here } a=1, b=a, c=-1$
$ \Rightarrow D=b^2-4 a c$
$ \Rightarrow D=(a)^2-4 \times 1 \times(-1)=a^2+4$
Roots are real,
$\Rightarrow\text{D}\geq0$
$\Rightarrow\text{a}^2+4\geq0$
For all real values of a, the equation has real roots.
View full question & answer→Question 1053 Marks
Write the number of real roots of the equation $ x^2+3|x|+2=0 $
Answer$ x^2+3|x|+2=0 $
$ x^2+3 x+2=0$
$\text { Here } \mathrm{x}=1, \mathrm{~b}=3, \mathrm{c}=2(\because|\mathrm{x}|=\mathrm{x})$
$ D=b^2-4 a c $
$ D=(3)^2-4 \times 1 \times 2 $
$ D=9-8 $
$ D=1$
$\text{x} = {-\text{b} \pm \sqrt{\text{b}^2-4\text{ac}} \over 2\text{a}}$
$\text{x}=\frac{-3\pm\sqrt{1}}{2\times1}$
$\text{x}=\frac{-3\pm{1}}{2}$
$\text{x}_1=\frac{-3+1}{2}$
$\text{x}_1=\frac{-2}{2}$
$\text{x}_1=-1$
and $\text{x}_2=\frac{-3-1}{2}$
$\text{x}_2=\frac{-4}{2}$
$\text{x}_2=-2$
$\therefore$ Real roots are $-1, -2$
View full question & answer→Question 1063 Marks
Find the value of k for which the following equation have real root:
$x^2-4 k x+k=0$
Answer$x^2-4 k x+k=0$
Here, comparing with $\mathrm{ax}^2+\mathrm{kx}+\mathrm{c}=0$
$a=1, b=-4 k, c=k $
$ D=b^2-4 a c $
$ D=(-4 k)^2-4 \times 1 \times k $
$ D=16 k^2-4 k$
$\because$ Roots are real and equal
$ \therefore D=0 $
$ \therefore 16 k^2-4 k=0 $
$ \Rightarrow 4 k(4 k-1)=0 $
$ \Rightarrow k(4 k-1)=0$
Either $\mathrm{k}=0$
or $4 \mathrm{k}-1=0$
$4 \mathrm{k}=1 $
$ \therefore \mathrm{k}=\frac{1}{4}$
Hence $k=0, \frac{1}{4}$
View full question & answer→Question 1073 Marks
If $1+\sqrt{2}$ is a root of a quadratic equation will rational coefficients, write its other root.
AnswerGiven that $(1+\sqrt{2})$ is a root of the quadratic equation with rational coefficients.
Then find the other root.
As we know that if $(1+\sqrt{2})$ is a root of the quadratic equation with rational.
coefficients then other roots be $(1+\sqrt{2})$
Hence, the require root of the quadratic equation be $(1+\sqrt{2})$
View full question & answer→Question 1083 Marks
A car moves a distance of $2592\ km$ with uniform speed. The number of hours taken for the journey is one-half the number representing the speed, in km/hour. Find the time taken to cover the distance.
AnswerDistance $= 2592\ km$
Let the speed of the car $= x\ km/hr$
and time taken $=\frac{\text{x}}{2}\text{hr}$
We have, Distance $=$ Speed $×$ Time
$\Rightarrow2592=\text{x}\times\frac{\text{x}}{2}$
$\Rightarrow2592=\frac{\text{x}^2}{2}$
$\Rightarrow\text{x}^2=2592\times2$
$\Rightarrow\text{x}=\sqrt{5184}$
$\Rightarrow\text{x}=72\text{km/hr}$
and thus time taken $=\frac{\text{x}}{2}\text{h}=\frac{72}{2}=36\text{ hr}$
View full question & answer→Question 1093 Marks
Find two consecutive natural numbers whose product is $20.$
AnswerLet the two consecutive natural numbers be $ 'x'$ and $'x + 1'$
Given that product of the natural numbers is $20$
$ \text { Hence, } x(x+1)=20$
$\Rightarrow x^2+x=20$
$\Rightarrow x^2+x-20=0$
$\Rightarrow x^2+5 x-4 x-20=0$
$\Rightarrow x(x+5)-4(x+5)=0$
$\Rightarrow x=-5 \text { or } x=4$
Considering positive value of x as $\text{x}\in\text{N}$
For $r = 4, x + 1 = 4 + 1 = 5$
The two consecutive natural numbers are $4$ as $5$
View full question & answer→Question 1103 Marks
Write a quadratic polynomial, sum of whose zeros is $2\sqrt{3}$ and their product is $2.$
AnswerAs we know that the quadratic
polynomial $f(x) = k[x^2- ($sum of their roots$)x + ($product of their roots$)]$
According to question,
$($sum of their roots$) =2\sqrt{3}$
And $($product of their roots$) = 2$
Thus Putting the value in above,
$\text{f}(\text{x})=\text{k}\big[\text{x}^2-2\sqrt{3}\text{x}+2\big]$ where k is real number.
Therefore, the quadratic polynomial be
$\text{f}(\text{x})=\text{k}\big[\text{x}^2-2\sqrt{3}\text{x}+2\big]$
View full question & answer→Question 1113 Marks
The difference of the squares of two positive integers is $180.$ The square of the smaller number is $8$ times the larger, find the numbers.
AnswerLet the larger number be $x$
Then according to the question,
$ \Rightarrow x^2-8 x=180 $
$ \Rightarrow x^2-8 x-180=0 $
$ \Rightarrow x^2-18 x+10 x-180=0$
$ \Rightarrow x(x-18)+10(x-18)=0$
$ \Rightarrow(x+10)(x-18)=0$
$ \Rightarrow x+10=0 \text { or } x-18=0 $
$ \Rightarrow x=-10 \text { or } x=18$
Since, x being a positive integer so, $x$ cannot be negative,
Therefore, larger number $= 18$
Then the smaller number $=\sqrt{8\times18}=12$
Thus, the two positive number are $12$ and $18$
View full question & answer→Question 1123 Marks
There are three consecutive integers such that the square of the first increased by the product of the other two gives $154.$ What are the integers.
AnswerLet first integer $= x$
Then second integer $= x + 1,$ and third integer $= x + 2$
According to the condition,
$ \Rightarrow x^2+(x+1)(x+2)=154 $
$ \Rightarrow x^2+x^2+3 x+2=154 $
$ \Rightarrow 2 x^2+3 x+2-154=0 $
$ \Rightarrow 2 x^2-16 x+19 x-152=0 $
$ \Rightarrow 2 x(x-8)+19(x-8)=0 $
$ \Rightarrow(x-8)(2 x+9)=0$
Either $x - 8 = 0,$ then $x = 8$
or $2x + 19 = 0$, then $2x = -19$
$\Rightarrow\text{x}=\frac{-19}{2}$ But it is not an integer.
First number $= 8$
Second number $= 8 + 1 = 9$ and third number $= 8 + 2 = 10$
View full question & answer→Question 1133 Marks
In the following, determine whether the given quadratic equation have real root and if so, find the root:
$\sqrt{2}\text{x}^2+7\text{x}+5\sqrt{2}=0$
Answer$\sqrt{2}\text{x}^2+7\text{x}+5\sqrt{2}=0$
The given equation is in the form of $\text{ax}^2+\text{bx}+\text{c}=0$
Here, $\text{a}=\sqrt{2},\text{b}=7,\text{c}=5\sqrt{2}$
The discriminant $\text{D}=\text{b}^2-4\text{ac}$
$=(7)^2-4\times\sqrt{2}\times5\sqrt{2}$
$=49-40$
$=9>0$
As $0 = 0,$ the given equation has roots real given by
$\Rightarrow\alpha=-\frac{\text{b}+\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-7+\sqrt{9}}{2\times\sqrt{2}}=\frac{-7+3}{2\sqrt{2}}$
$=\frac{-\text{4}}{2\sqrt{2}}=-\sqrt{2}$
$\Rightarrow\beta=-\frac{\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-7-\sqrt{9}}{2\sqrt{2}}=\frac{-7-3}{2\sqrt{2}}$
$=\frac{-10}{2\sqrt{2}}=\frac{-5}{\sqrt{2}}$
$\therefore$ The roots of the given equation are $-\sqrt{2},\frac{-5}{\sqrt{2}}$
View full question & answer→Question 1143 Marks
Determine the nature of the root of following quadratic equation:
$(b + c)x^2- (a + b + c)x + a = 0$
AnswerThe given quadric equation is $(b+c) x^2-(a+b+c) x+a=0$
Here, $a=(b+c), b=-(a+b+c)$ and $c=a$
As we know that $D=b^2-4 a c$
Putting the value of $a=(b+c), b=-(a+b+c)$ and $c=a$
$=(-(a+b+c))^2-4 \times(b+c) \times a $
$=\left(a^2+b^2+c^2+2 a b+2 b c+2 c a\right)-4 a b-4 c a$
$ =a^2+b^2+c^2-2 a b+2 b c-2 c a$
Since, $D > 0$
Therefore, root of the given equation are real and unequal.
View full question & answer→Question 1153 Marks
In the following, determine whether the given values are solution of the given equation or not:
$\text{x}^2-3\sqrt{3}\text{x}+6=0,$ $\text{x}=\sqrt{3},\text{x}=-2\sqrt{3}$
AnswerWe have been given that,
$\text{x}^2-3\sqrt{3}\text{x}+6=0,$ $\text{x}=\sqrt{3},\text{x}=-2\sqrt{3}$
Now, if $\text{x}=\sqrt{3}$ is a solution then it should satisfy the equation so, substituting $\text{x}=\sqrt{3}$ in the equation, we get
$\text{x}^2-3\sqrt{3}\text{x}+6=(\sqrt{3})^2-3\sqrt{3}(\sqrt{3})+6$
$=3-9+6$
$=0$
$\text{x}=-2\sqrt{3}$
$(-2\sqrt{3})^2-3\sqrt{3}(-2\sqrt{3})+6$
$4\times3+6\times3+6$
$=12+18+6$
$=36$
Which in not equal to zero.
Hence, $\text{x}=-2\sqrt{3}$ is a solution of the quadratic equation.
Therefore, from the above results we find out that $\text{x}=\sqrt{3}$ and $\text{x}=-2\sqrt{3}$ is not solutions of the given quadratic equation.
View full question & answer→Question 1163 Marks
Determine the set of values of k for which the given following quadratic has real root:
$2x^2+ kx + 2 = 0$
AnswerThe given quadric equation is $2 x^2+k x+2=0$, and roots are real.
Then find the value of $k$.
Here, $a=2, b=k$ and $c=2$
As we know that of $a=2, b=k$ and $c=2$
$=(k)^2-4 \times 2 \times 2 $
$ =k^2-16$
The given equation will have real roots, if $\text{D}\geq0$
$\text{k}^2-16\geq0$
$\text{k}^2\geq16$
$\text{k}\geq\sqrt{16}$ or $\text{k}\leq-\sqrt{16}$
$\text{k}\leq-4$ or $\text{k}\geq4$
Therefore, the value of $\text{k}\leq-4$ or $\text{k}\geq4$
View full question & answer→Question 1173 Marks
Find the value of k for which the root are real and equal in the following equations:
$3 x^2-5 x+2 k=0$
AnswerThe given equation is $3 x^2-5 x+2 k=0$
This equation is in the from of $a x^2+b x+c=0$
Here, $a=3, b=-5$ and $c=2 k$
Given that, the equation has real and equal roots
$\text { i.e., } D=b^2-4 a c=0$
$ \Rightarrow(-5)^2-4 \times 3 \times(2 \mathrm{k})=0$
$ \Rightarrow 25=24 \mathrm{k}$
$\Rightarrow\text{k}=\frac{25}{24}$
$\therefore$ The value of $\text{k}=\frac{25}{24}$
View full question & answer→Question 1183 Marks
Find the roots of the quadratic equation:
$\sqrt{2}\text{x}^2+7\text{x}+5\sqrt{2}=0$
Answer$\sqrt{2}\text{x}^2+7\text{x}+5\sqrt{2}=0$
$\Rightarrow\sqrt{2}\text{x}^2+5\text{x}+2\text{x}+5\sqrt{2}=0$
$\Rightarrow\text{x}\big(\sqrt{2}\text{x}+5\big)+\sqrt{2}\big(\sqrt{2}\text{x}+5\big)=0$
$\Rightarrow\big(\sqrt{2}\text{x}+5\big)\big(\text{x}+\sqrt{2}\big)=0$
Either $\sqrt{2}\text{x}+5=0,$ then $\text{x}=\frac{-5}{\sqrt{2}}$
or $\text{x}+\sqrt{2}=0,$ then $\text{x}=-\sqrt{2}$
$\therefore$ Roots are $\frac{-5}{\sqrt{2}},-\sqrt{2}$
View full question & answer→Question 1193 Marks
The difference of two numbers is $4.$ If the difference of their reciprocal is $\frac{4}{21},$ find the numbers.
AnswerLet the two numbers be $x$ and $x - 4$
Given that the difference of two numbers is $4$
By the given hypothesis, we have $\frac{1}{\text{x}-4}-\frac{1}{\text{x}}=\frac{4}{21}$
$\Rightarrow\frac{\text{x}-\text{x}+4}{\text{x}(\text{x}-4)}=\frac{4}{21}$
$ \Rightarrow 84=4 x(x-4) $
$ \Rightarrow x^2-4 x-21=0 $
$ \Rightarrow x^2-7 x+3 x-21=0 $
$ \Rightarrow x(x-7)+3(x-7)=0 $
$ \Rightarrow(x-7)(x+3)=0$
$⇒ x = 7$ or $ x = -3 $ and
If $x = -3, x - 4 = -3 - 4 = -7$
Hence, required numbers are $3, 7$ and $-3, -7x$
View full question & answer→Question 1203 Marks
A train travels $360\ km$ at a uniform speed. If the speed had been $5\ km/hr$ more, it would have taken $1$ hour less for the same journey. Form the quadratic equation to find the speed of the train.
AnswerTotal distance $= 360km$
Let the uniform speed of the train $= x km/hr$
Time taken $=\frac{360}{\text{x}}$
In second case speed $= (x + 5) km/hr$
$\therefore$ Time taken $=\frac{360}{\text{x}+5}\text{h}$
$\therefore\frac{360}{\text{x}}-\frac{360}{\text{x}+5}=1$
$⇒ 360x + 1800 - 360x = x(x + 5)$
$\Rightarrow 1800=x^2+5 x \Rightarrow x^2+5 x-1800=0$
$\therefore \text { Required quadratic equation will be }$
$x^2+5 x-1800=0$
View full question & answer→Question 1213 Marks
Solve the following quadratic equations by factorization:
$16\text{x}-\frac{10}{\text{x}}=27$
Answer$16\text{x}-\frac{10}{\text{x}}=27$
$16 x^2-10=27 x$
$16 x^2-27 x-10=0$
$16 x^2-32 x+5 x-10=0$
$16 x(x-2)+5(x-2)=0$
$(16 x+5)(x-2)=0$
$16 x+5=0 \text { or } x-2=0$
$\Rightarrow\text{x}=-\frac{5}{16}$ or $x = 2$
Hence, the factors are $2$ and $-\frac{5}{16}$
View full question & answer→Question 1223 Marks
Find the roots of the following quadratic equation (if they exist) by the method of completing the square.
$x^2-4 a x+4 a^2-b^2=0$
Answer$\text{x}^2-4\text{ax}+4\text{a}^2-\text{b}^2=0$
$\Rightarrow\text{x}^2-2\cdot2\text{a}\cdot\text{x}+(2\text{a})^2=\text{b}^2$
$\Rightarrow(\text{x}-2\text{a})^2=(\pm\text{b})^2$
$\Rightarrow\text{x}-2\text{a}=\pm\text{b}$
$\therefore\text{x}=2\text{a}\pm\text{b}$
$\therefore\text{x}=2\text{a}+\text{b}$
and $\text{x}=2\text{a}-\text{b}$
Hence roots are $2\text{a}+\text{b},2\text{a}-\text{b}$
View full question & answer→Question 1233 Marks
Solve the following quadratic equation by factorization:
$ a^2 b^2 x^2+b^2 x-a^2 x-1=0$
AnswerWe have
$ a^2 b^2 x^2+b^2 x-a^2 x-1=0$
${\left[-1 \times a^2 b^2=-a^2 b^2 \Rightarrow-a^2 b^2=-a^2 \times b^2=-a^2 \times b^2\right]}$
$\Rightarrow a^2 b^2 x^2+b^2 x-a^2 x-1=0$
$\Rightarrow b^2 x\left(a^2 x+1\right)-1\left(a^2 x+1\right)=0$
$\Rightarrow\left(a^2 x+1\right)\left(b^2 x-1\right)=0$
$\Rightarrow a^2 x+1=0 \text { or } b^2 x-1=0$
$\Rightarrow\text{x}= -\frac{1}{\text{a}^2}$ and $\text{x}= \frac{1}{\text{b}^2}$ are the two root of the given equation.
View full question & answer→Question 1243 Marks
In the following, determine the set of values of k for which the given quadratic equation has real root:
$2x^2+ x + k = 0$
AnswerThe given equation is $2x^2+ x + k = 0$, and roots are real.
Then find the value of k.
Here, $a = 2, b = 1, c = k$
As we know that $D = b^2- 4ac$
Putting the value of $a = 2, b = 1, c = k$
$D = 1 - 8k$
The given equation will have real roots, if $\text{D}\geq0$
$\text{D}=1-8\text{k}\geq0$
$\Rightarrow8\text{k}\leq1$
$\Rightarrow\text{k}\leq\frac{1}{8}$
Therefore, the value of $\text{k}\leq\frac{1}{8}$
View full question & answer→Question 1253 Marks
Determine the nature of the root of following quadratic equation:
$ 2\left(a^2+b^2\right) x^2+2(a+b) x+1=0$
Answer$ 2\left(a^2+b^2\right) x^2+2(a+b) x+1=0$
$\text { Here, } a=2\left(a^2+b^2\right), b=2(a+b), c=1$
$\therefore \text { Discriminant }(D)=b^2-4 a c$
$=\{2(a+b)\}^2-4 \times 2\left(a^2+b^2\right) \times 1$
$=4\left(a^2+b^2+2 a b\right)-8\left(a^2+b^2\right)$
$=4 a^2+4 b^2+8 a b-8 a^2-8 b^2$
$=8 a b-4 a^2-4 b^2$
$=-\left(4 a^2+4 b^2-8 a b\right)$
$=-4\left(a^2+b^2-2 a b\right)$
$=-4(a-b)^2$
$\because D<0$
$\therefore$ Roots are not real.
View full question & answer→Question 1263 Marks
The sum of the squares of two consecutive odd numbers is $394.$ Find the numbers.
AnswerLet the consecutive odd positive integers are $2x - 1$ and $2x + 1$
Given that the sum of the squares is 394
$ \Rightarrow(2 x-1)^2+(2 x+1)^2=394 $
$ \Rightarrow 4 x^2+1-4 x+4 x^2+1+4 x=394 $
$ \Rightarrow 8 x^2+2=394 $
$ \Rightarrow 8 x^2=394-2 $
$ \Rightarrow 8 x^2=392$
$\Rightarrow\text{x}^2=\frac{392}{8}$
$x = 7$
As $x = 7, 2x - 1 = 2 × 7 - 1 = 14 - 1 = 13$
$⇒ 2x + 1 = 2 × 7 + 1 = 14 + 1 = 15$
The two consecutive odd positive numbers are $11$ and $13.$
View full question & answer→Question 1273 Marks
Find the roots of the following quadratic equation (if they exist) by the method of completing the square.
$\sqrt{2}\text{x}^2-3\text{x}-2\sqrt{2}=0$
Answer $\sqrt{2}\text{x}^2-3\text{x}-2\sqrt{2}=0$
$\Rightarrow\text{x}^2-\frac{3}{\sqrt{2}}\text{x}-2=0$
$($Dividng by $\sqrt{2})$
$\Rightarrow(\text{x}^2)-2\times\frac{3}{2\sqrt{2}}\text{x}+\Big(\frac{3}{2\sqrt{2}}\Big)^2-\frac{25}{8}=0$
$\Rightarrow\Big(\text{x}-\frac{3}{2\sqrt{2}}\Big)^2=\Big(\pm\frac{5}{2\sqrt{2}}\Big)^2$
$\begin{cases}\because-2=\frac{9}{8}-\frac{25}{8}\\=\frac{3}{2\sqrt{2}}-\frac{25}{8}\end{cases}$
$\therefore\text{x}-\frac{3}{2\sqrt{2}}=\pm\frac{5}{2\sqrt{2}}$
$\Rightarrow\text{x}=\frac{3}{2\sqrt{2}}\pm\frac{5}{2\sqrt{2}}$
$=\frac{3}{2\sqrt{2}}\pm\frac{5}{2\sqrt{2}}$
$=\frac{8}{2\sqrt{2}}=\frac{4}{\sqrt{2}}=2\sqrt{2}$
and $\text{x}=\frac{3}{2\sqrt{2}}-\frac{5}{2\sqrt{2}}$
$=\frac{-2}{2\sqrt{2}}=\frac{-1}{\sqrt{2}}$
Roots are $=2\sqrt{2}$ and $\frac{-1}{\sqrt{2}}$
View full question & answer→Question 1283 Marks
In the following, determine whether the given quadratic equation have real root and if so, find the root:
$x^2-2 x+1=0$
Answer$x^2-2 x+1=0$
The given equation is in the form of $a x^2+b x+c=0$
Here $\mathrm{a}=1, \mathrm{~b}=-2$ and $\mathrm{c}=1$
The discriminant $D=b^2-4 a c$
$\Rightarrow(-2)^2-4 \times 1 \times 1=0$
As $Q = 0,$ the given equation has real and equal roots
$\Rightarrow\alpha=-\frac{\text{b}+\sqrt{\text{D}}}{2\text{a}},$ $\beta=-\frac{\text{b}-\sqrt{\text{D}}}{2\text{a}}$
i.e., $\alpha$ and $\beta=-\frac{\text{b}}{2\text{a}}\ [\because0=0]$
$\Rightarrow\alpha$ and $\beta=-\frac{\text{b}}{2\text{a}}$
$=-\frac{(-2)}{2\times2}=\frac{2}{2}=1$
The roots of the given equation $\alpha$ and $\beta$ is $1$
View full question & answer→Question 1293 Marks
In the following, determine whether the given quadratic equation have real root and if so, find the root:
$2\text{x}^2+5\sqrt{3}\text{x}+6=0$
Answer$2\text{x}^2+5\sqrt{3}\text{x}+6=0$
Here, $\text{a}=2,\text{b}=5\sqrt{3},\text{c}=6$
$\therefore$ Discriminant $(\text{D})=\text{b}^2-4\text{ac}$
$=(5\sqrt{3})^2-4\times2\times6$
$=75-48=27$
$\because\text{D}>0$
$\therefore$ Roots are real and unequal
$\therefore\text{x} = {-\text{b} \pm \sqrt{\text{b}^2-4\text{ac}} \over 2\text{a}}$
$=\frac{-5\sqrt{3}\pm\sqrt{27}}{2\times2}$
$=\frac{-5\sqrt{3}\pm3\sqrt{3}}{4}$
$\therefore\text{x}=\frac{-5\sqrt{3}+3\sqrt{3}}{4}=\frac{-2\sqrt{3}}{4}$
$\text{x}=\frac{-\sqrt{3}}{2}$
and $\text{x}=\frac{-5\sqrt{3}-3\sqrt{3}}{4}=\frac{-8\sqrt{3}}{4}$
$\text{x}=-2\sqrt{3}$
Roots are $\frac{-\sqrt{3}}{2},-2\sqrt{3}$
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