Question

Find whether the following equation have real roots. If real roots exist, find them
(i) $8x^2+2x-3=0$
(ii) $-2x^2+3x+2=0$
(iii) $5x^2-2x-10=0$
(iv) $\frac{1}{2x-3}+\frac{1}{x-5}=1,x\ne \frac{3}{2},5$
(v) $x^2+5\sqrt{5}x-70=0$

Answer 1

(i) Given equation is $8x^2+2x-3=0$
On comparing with $a{x}^2+bx+c=0$, we get:
$a=8,b=2\text{and}c=-3$
$\therefore \text{Discriminant,}D=b^2-4ac$
$=(2{)}^2-4(8\left)\right(-3)$
$=4+96$
$=100>0$
As D > 0, the equation $8x^2+2x-3=0$ has two distinct real roots.

We know that, Quadratic formula
$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$x=\frac{-2\pm \sqrt{4+96}}{16}$
$=\frac{-2\pm \sqrt{100}}{16}=\frac{-2\pm 10}{16}$
$\Rightarrow x=\frac{1}{2}orx=-\frac{3}{4}$
Hence, the roots of the equation
$8x^2+2x-3=0\text{are}\frac{1}{2}\text{and}-\frac{3}{4}$

(ii) Given equation is $-2x^2+3x+2=0$
On comparing with $a{x}^2+bx+c=0,$ we get $a=-2,b=3\text{and}c=2$
$\therefore \text{Discriminant,}D=b^2-4ac$
$=(3{)}^2-4(-2\left)\right(2)$
$=9+4\times 4=9+16=25>0$
As D > 0, the equation $-2x^2+3x+2=0$ has two distinct real roots.
We know that, Quadratic formula is
$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$=\frac{-3\pm \sqrt{(3{)}^2-4(-2\left)\right(2)}}{2(-2)}=\frac{-3\pm \sqrt{9+16}}{-4}$
$=\frac{-3\pm \sqrt{25}}{-4}=\frac{-3\pm 5}{-4}$
$\Rightarrow x=\frac{-3+5}{-4}=-\frac{1}{2}orx=\frac{-3-5}{-4}=\frac{-8}{-4}=2$
Hence, the roots of the equation
$-2x^2+3x+2=0\text{are}(-\frac{1}{2})\text{and}2.$

(iii) Given equation is $5x^2-2x-10=0$
On comparing with $a{x}^2+bx+c=0,$ we get: $a=5,b=-2\text{and}c=-10$
Discriminant, $D=b^2-4ac$
$=(-2{)}^2-4(5\left)\right(-10)$
$=4+200=204>0$
As D > 0, the equation $5x^2-2x-10=0$ has two distinct real roots.
We know that, Quadratic formula is
$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$=\frac{-(-2)\pm \sqrt{(-2{)}^2-4(5\left)\right(-10)}}{2\times 5}$
$=\frac{2\pm \sqrt{4+200}}{10}=\frac{2\pm \sqrt{204}}{10}=\frac{2\pm \sqrt{51\times 4}}{10}$
$=\frac{2\pm 2\sqrt{51}}{10}=\frac{1\pm \sqrt{51}}{5}$
$\Rightarrow x=\frac{1+\sqrt{51}}{5}orx=\frac{1-\sqrt{51}}{5}$
Hence, the roots of the equation
$5x^2-2x-10=0\text{are}\frac{1+\sqrt{51}}{5}\text{and}\frac{1-\sqrt{51}}{5}$

(iv) Given equation is $\frac{1}{2x-3}+\frac{1}{x-5}=1,x\ne \frac{3}{2},5$
$\Rightarrow \frac{x-5+2x-3}{(2x-3)(x-5)}=1$
$\Rightarrow \frac{3x-8}{2x^3-10x-3x+15}=1$
$\Rightarrow 3x-8=2x^2-10x-3x+15$
$\Rightarrow 2x^2-16x+23=0$
On comparing with $a{x}^2+bx+c=0$ we get: $a=2,b=-16\text{and}c=23$
We know that Discriminant $\left(D\right)=b^2-4ac$
$=(-16{)}^2-4(2\left)\right(23)=256-184=72>0$
As, D > 0 the equation $2x^2-16x+23=0$ has two distinct real roots.
We know that, Quadratic formula
$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$=\frac{-(-16)\pm \sqrt{(-16{)}^2-4(2\left)\right(23)}}{2\times 2}$
$=\frac{16\pm \sqrt{256-184}}{4}$
$=\frac{16\pm \sqrt{72}}{4}=\frac{16\pm 6\sqrt{2}}{4}=\frac{8\pm 3\sqrt{2}}{4}$
$\Rightarrow x=\frac{8+3\sqrt{2}}{2}orx=\frac{8-3\sqrt{2}}{2}$
Hence, the roots of the equation
$\frac{1}{2x-3}+\frac{1}{x-5}=1\text{are}\frac{8+3\sqrt{2}}{2}\text{and}\frac{8-3\sqrt{2}}{2}$

(v) Given equation $x^2+5\sqrt{5}x-70=0$
On comparing with $a{x}^2+bx+c=0,$ we get: $a=1,b=5\sqrt{5}\text{and}c=-70$
We know that Discriminant $\left(D\right)=b^2-4ac$
$=(5\sqrt{5}{)}^2-4(1\left)\right(-70)=125+280$
$=405>0$
As D > 0, the equation $x^2+5\sqrt{5}x-70=0$ has two distinct real roots.
We know that, Quadratic formula
$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$=\frac{-5\sqrt{5}\pm \sqrt{(5\sqrt{5}{)}^2-4(1\left)\right(-70)}}{2\times 1}$
$=\frac{-5\sqrt{5}\pm \sqrt{125+280}}{2\times 1}=\frac{-5\sqrt{5}\pm \sqrt{405}}{2}$
$=\frac{-5\sqrt{5}\pm 9\sqrt{5}}{2}$
$\Rightarrow x=\frac{-5\sqrt{5}+9\sqrt{5}}{2}orx=\frac{-5\sqrt{5}-9\sqrt{5}}{2}$
$\Rightarrow x=\frac{4\sqrt{5}}{2}orx=\frac{-14\sqrt{5}}{2}$
Hence, the roots of the equation $x^2+5\sqrt{5}x-70=0\text{are}2\sqrt{5}\text{and}-7\sqrt{5}.$