Find the roots of the quadratic equations by using the quadratic formula in each of the following:
$ 2 x^{2}-3 x-5=0 $

AcademicMathematicsNCERTClass 10

To do:

We have to find the roots of the given quadratic equations.

Solution:

(i) $2x^2-3x - 5 = 0$

The above equation is of the form $ax^2 + bx + c = 0$, where $a = 2, b = -3$ and $c = - 5$

Discriminant $\mathrm{D} =b^{2}-4 a c$

$=(-3)^{2}-4 \times 2(-5)$

$=9+40$

$=49$

$\mathrm{D}>0$

Let the roots of the equation are $\alpha$ and $\beta$

$\alpha =\frac{-b+\sqrt{\mathrm{D}}}{2 a}$

$=\frac{-(-3)+\sqrt{49}}{2(2)}$

$=\frac{3+7}{4}$

$=\frac{10}{4}$

$=\frac{5}{2}$

$\beta =\frac{-b-\sqrt{\mathrm{D}}}{2 a}$

$=\frac{-(-3)-\sqrt{49}}{2(2)}$

$=\frac{3-7}{4}$

$=\frac{-4}{4}$

$=-1$

Hence, the roots of the given quadratic equation are $\frac{5}{2}, -1$. 

(ii) $5x^2+13x + 8 = 0$

The above equation is of the form $ax^2 + bx + c = 0$, where $a = 5, b = 13$ and $c =  8$

Discriminant $\mathrm{D} =b^{2}-4 a c$

$=(13)^{2}-4 \times 5\times8$

$=169-160$

$=9$

$\mathrm{D}>0$

Let the roots of the equation are $\alpha$ and $\beta$

$\alpha =\frac{-b+\sqrt{\mathrm{D}}}{2 a}$

$=\frac{-13+\sqrt{9}}{2(5)}$

$=\frac{-13+3}{10}$

$=\frac{-10}{10}$

$=-1$

$\beta =\frac{-b-\sqrt{\mathrm{D}}}{2 a}$

$=\frac{-13-\sqrt{9}}{2(5)}$

$=\frac{-13-3}{10}$

$=\frac{-16}{10}$

$=-\frac{8}{5}$

Hence, the roots of the given quadratic equation are $-\frac{8}{5}, -1$. 

(iii) $-3x^2+5x + 12 = 0$

The above equation is of the form $ax^2 + bx + c = 0$, where $a = -3, b = 5$ and $c =  12$

Discriminant $\mathrm{D} =b^{2}-4 a c$

$=(5)^{2}-4 \times (-3)\times12$

$=25+144$

$=169$

$\mathrm{D}>0$

Let the roots of the equation are $\alpha$ and $\beta$

$\alpha =\frac{-b+\sqrt{\mathrm{D}}}{2 a}$

$=\frac{-5+\sqrt{169}}{2(-3)}$

$=\frac{-5+13}{-6}$

$=\frac{8}{-6}$

$=-\frac{4}{3}$

$\beta =\frac{-b-\sqrt{\mathrm{D}}}{2 a}$

$=\frac{-5-\sqrt{169}}{2(-3)}$

$=\frac{-5-13}{-6}$

$=\frac{-18}{-6}$

$=3$

Hence, the roots of the given quadratic equation are $-\frac{4}{3}, 3$. 

(iv) $-x^2+7x - 10 = 0$

The above equation is of the form $ax^2 + bx + c = 0$, where $a = -1, b = 7$ and $c =-10$

Discriminant $\mathrm{D} =b^{2}-4 a c$

$=(7)^{2}-4 \times (-1)\times(-10)$

$=49-40$

$=9$

$\mathrm{D}>0$

Let the roots of the equation are $\alpha$ and $\beta$

$\alpha =\frac{-b+\sqrt{\mathrm{D}}}{2 a}$

$=\frac{-7+\sqrt{9}}{2(-1)}$

$=\frac{-7+3}{-2}$

$=\frac{-4}{-2}$

$=2$

$\beta =\frac{-b-\sqrt{\mathrm{D}}}{2 a}$

$=\frac{-7-\sqrt{9}}{2(-1)}$

$=\frac{-7-3}{-2}$

$=\frac{-10}{-2}$

$=5$

Hence, the roots of the given quadratic equation are $2, 5$. 

(v) \( x^{2}+2 \sqrt{2} x-6=0 \)

The above equation is of the form $ax^2 + bx + c = 0$, where $a = 1, b = 2 \sqrt{2}$ and $c =-6$

Discriminant $\mathrm{D} =b^{2}-4 a c$

$=(2 \sqrt{2})^{2}-4 \times (1)\times(-6)$

$=8+24$

$=32$

$\mathrm{D}>0$

Let the roots of the equation are $\alpha$ and $\beta$

$\alpha =\frac{-b+\sqrt{\mathrm{D}}}{2 a}$

$=\frac{-2 \sqrt{2}+\sqrt{32}}{2(1)}$

$=\frac{-2 \sqrt{2}+4\sqrt2}{2}$

$=\frac{2\sqrt2}{2}$

$=\sqrt2$

$\beta =\frac{-b-\sqrt{\mathrm{D}}}{2 a}$

$=\frac{-2 \sqrt{2}-\sqrt{32}}{2(1)}$

$=\frac{-2 \sqrt{2}-4\sqrt{2}}{2}$

$=\frac{-6\sqrt{2}}{2}$

$=-3\sqrt2$

Hence, the roots of the given quadratic equation are $\sqrt2, -3\sqrt2$. 

(vi) We know that for a quadratic equation $ax^{2} +bx+c=0$

 $x=\frac{-b\pm \sqrt{b^{2} -4ac}}{2a}$

On comparing it to the given quadratic equation $a=1,b=-3\sqrt{5} \ and\ c=10$

On substituting these values of $\displaystyle a,\ b\ and\ c$

$x=\frac{-( -3\surd 5) \pm \sqrt{( -3\surd 5)^{2} -4\times 1\times 10}}{2\times 1}$

$x=\frac{3\sqrt{5} \pm \sqrt{( 45-40)}}{2}$

$x=\frac{\left( 3\sqrt{5} \pm \sqrt{5}\right)}{2}$

If $x=\frac{\left( 3\sqrt{5} +\sqrt{5}\right)}{2}$

$\Rightarrow x=\frac{4\sqrt{5}}{2} $

$\Rightarrow x=2\sqrt{5}$

If $x=\frac{\left( 3\sqrt{5} -\sqrt{5}\right)}{2}$

$\Rightarrow x=\frac{\left( 2\sqrt{5}\right)}{2}$

$\Rightarrow x=\sqrt{5}$

$\therefore x=2\sqrt{5}, \ \sqrt{5}$

(vii) \( \frac{1}{2} x^{2}-\sqrt{11} x+1=0 \)

The above equation is of the form $ax^2 + bx + c = 0$, where $a = \frac{1}{2}, b = -\sqrt{11}$ and $c =1$

Discriminant $\mathrm{D} =b^{2}-4 a c$

$=(-\sqrt{11})^{2}-4 \times \frac{1}{2} \times 1$

$=11-2$

$=9$

$\mathrm{D}>0$

Let the roots of the equation are $\alpha$ and $\beta$

$\alpha =\frac{-b+\sqrt{\mathrm{D}}}{2 a}$

$=\frac{-(-\sqrt{11})+\sqrt{9}}{2(\frac{1}{2})}$

$=\frac{\sqrt{11}+3}{1}$

$=3+\sqrt{11}$

$\beta =\frac{-b-\sqrt{\mathrm{D}}}{2 a}$

$=\frac{-(-\sqrt{11})-\sqrt{9}}{2(\frac{1}{2})}$

$=\frac{\sqrt{11}-3}{1}$

$=-3+\sqrt{11}$

Hence, the roots of the given quadratic equation are $3+\sqrt{11}, -3+\sqrt{11}$. 

raja
Updated on 10-Oct-2022 13:27:26

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