Find whether the following equations have real roots. If real roots exist, find them.
$ 8 x^{2}+2 x-3=0 $

AcademicMathematicsNCERTClass 10

To do:

We have to determine whether the given quadratic equations have real roots.

Solution:

(i) Comparing the given quadratic equation with the standard form of the quadratic equation $ax^2+bx+c=0$, we get,

$a=8, b=2$ and $c=-3$.

The discriminant of the standard form of the quadratic equation $ax^2+bx+c=0$ is

$D=b^2-4ac$.

Therefore,

$D=(2)^2-4(8)(-3)=4+96=100$.

As $D>0$, the given quadratic equation has two distinct real roots.

This implies,

$x=\frac{-b\pm \sqrt{D}}{2a}$

$x=\frac{-2 \pm \sqrt{100}}{2(8)}$ 

$x=\frac{-2 \pm 10}{16}$ 

$x=\frac{-2+10}{16}$ or $x= \frac{-2-10}{16}$

$x=\frac{8}{16}$ or $x=\frac{-12}{16}$

$x=\frac{1}{2}$ or $x=\frac{-3}{4}$

The roots of the given quadratic equation are $\frac{1}{2}$ and $-\frac{3}{4}$. 

(ii) Comparing the given quadratic equation with the standard form of the quadratic equation $ax^2+bx+c=0$, we get,

$a=-2, b=3$ and $c=2$.

The discriminant of the standard form of the quadratic equation $ax^2+bx+c=0$ is

$D=b^2-4ac$.

Therefore,

$D=(3)^2-4(-2)(2)$

$=9+16$

$=25$.

As $D>0$, the given quadratic equation has two distinct real roots.

This implies,

$x=\frac{-b\pm \sqrt{D}}{2a}$

$x=\frac{-3 \pm \sqrt{25}}{2(-2)}$ 

$x=\frac{-3 \pm 5}{-4}$ 

$x=\frac{-3+5}{-4}$ or $x= \frac{-3-5}{-4}$

$x=\frac{2}{-4}$ or $x=\frac{-8}{-4}$

$x=-\frac{1}{2}$ or $x=2$

The roots of the given quadratic equation are $-\frac{1}{2}$ and $2$. 

(iii) Comparing the given quadratic equation with the standard form of the quadratic equation $ax^2+bx+c=0$, we get,

$a=5, b=-2$ and $c=-10$.

The discriminant of the standard form of the quadratic equation $ax^2+bx+c=0$ is

$D=b^2-4ac$.

Therefore,

$D=(-2)^2-4(5)(-10)$

$=4+200$

$=204$.

As $D>0$, the given quadratic equation has two distinct real roots.

This implies,

$x=\frac{-b\pm \sqrt{D}}{2a}$

$x=\frac{-(-2) \pm \sqrt{204}}{2(5)}$ 

$x=\frac{2 \pm 2\sqrt{51}}{10}$ 

$x=\frac{2(1+\sqrt{51})}{10}$ or $x= \frac{2(1-\sqrt{51})}{10}$

$x=\frac{1+\sqrt{51}}{5}$ or $x=\frac{1-\sqrt{51}}{5}$

The roots of the given quadratic equation are $\frac{1+\sqrt{51}}{5}$ and $\frac{1-\sqrt{51}}{5}$. 

(iv) $\frac{1}{2 x-3}+\frac{1}{x-5}=1$

$\frac{x-5+2 x-3}{(2 x-3)(x-5)}=1$

$\frac{3 x-8}{2 x^{2}-3x-10 x+15}=1$

$\frac{3 x-8}{2 x^{2}-13x+15}=1$

$3 x-8 =2 x^{2}-13 x+15$

$2 x^{2}-13 x-3 x+15+8 =0$

$2 x^{2}-16x+23=0$

Comparing the above quadratic equation with the standard form of the quadratic equation $ax^2+bx+c=0$, we get,

$a=2, b=-16$ and $c=23$.

The discriminant of the standard form of the quadratic equation $ax^2+bx+c=0$ is

$D=b^2-4ac$.

Therefore,

$D=(-16)^2-4(2)(23)$

$=256-184$

$=72$

As $D>0$, the given quadratic equation has two distinct real roots.

This implies,

$x=\frac{-b\pm \sqrt{D}}{2a}$

$x=\frac{-(-16) \pm \sqrt{72}}{2(2)}$ 

$x=\frac{16 \pm 6\sqrt{2}}{4}$ 

$x=\frac{2(8+3\sqrt{2})}{4}$ or $x= \frac{2(8-3\sqrt{2})}{4}$

$x=\frac{8+3\sqrt{2}}{2}$ or $x=\frac{8-3\sqrt{2}}{2}$

The roots of the given quadratic equation are $\frac{8+3\sqrt{2}}{2}$ and $\frac{8-3\sqrt{2}}{2}$. 

(v) Comparing the given quadratic equation with the standard form of the quadratic equation $ax^2+bx+c=0$, we get,

$a=1, b=5 \sqrt{5}$ and $c=-70$.

The discriminant of the standard form of the quadratic equation $ax^2+bx+c=0$ is

$D=b^2-4ac$.

Therefore,

$D=(5 \sqrt{5})^2-4(1)(-70)$

$=125+280$

$=405$.

As $D>0$, the given quadratic equation has two distinct real roots.

This implies,

$x=\frac{-b\pm \sqrt{D}}{2a}$

$x=\frac{-5 \sqrt{5} \pm \sqrt{405}}{2(1)}$ 

$x=\frac{-5 \sqrt{5}\pm 9\sqrt5}{2}$ 

$x=\frac{-5 \sqrt{5}+9 \sqrt{5}}{2}$ or $x= \frac{-5 \sqrt{5}-9 \sqrt{5}}{2}$

$x=\frac{4\sqrt{5}}{2}$ or $x=\frac{-14 \sqrt{5}}{2}$

$x=2\sqrt5$ or $x=-7\sqrt5$

The roots of the given quadratic equation are $-7\sqrt5$ and $2\sqrt5$. 

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

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