Find whether the following equations
Question:

Find whether the following equations have real roots. If real roots exist, find them

(i) 8x2 +2x -3=0

(ii) – 2x2 + 3x + 2 = 0

(iii) 5x2 – 2x- 10 = 0

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

(v) $x^{2}+5 \sqrt{5} x-70=0$

Solution:

(i) Given equation is $8 x^{2}+2 x-3=0$.

On comparing with $a x^{2}+b x+c=0$, we get

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

$\therefore \quad$ Discriminant, $D=b^{2}-4 a c$

$=(2)^{2}-4(8)(-3)$

$=4+96=100>0$

Therefore, the equation $8 x^{2}+2 x-3=0$ has two distinct real roots because we know that, if the equation $a x^{2}+b x+c=0$ has discriminant

greater than zero, then if has two distinct real roots.

 Roots,  $x=\frac{-b \pm \sqrt{D}}{2 a}=\frac{-2 \pm \sqrt{100}}{16}=\frac{-2 \pm 10}{16}$

$=\frac{-2+10}{16}, \frac{-1-10}{16}$

$=\frac{8}{16},-\frac{12}{16}=\frac{1}{2},-\frac{3}{4}$

(ii) Given equation is $-2 x^{2}+3 x+2=0$.

On comparing with $a x^{2}+b x+c=0$, we get

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

$\therefore \quad$ Discriminant, $D=b^{2}-4 a c$

$=(3)^{2}-4(-2)(2)$

$=9+16=25>0$

Therefore, the equation $-2 x^{2}+3 x+2=0$ has two distinct real roots because we know that if the equation $a x^{2}+b x+c=0$ has its discriminant greater than zero, then it has two distinct real roots.

Roots, $x=\frac{-b \pm \sqrt{D}}{2 a}=\frac{-3 \pm \sqrt{25}}{2(-2)}$

$=\frac{-3 \pm 5}{-4}=\frac{-3+5}{-4}, \frac{-3-5}{-4}$

$=\frac{2}{-4}, \frac{-8}{-4}=-\frac{1}{2}, 2$

(iii) Given equation is $5 x^{2}-2 x-10=0$.

On comparing with $a x^{2}+b x+c=0$, we get

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

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

$=(-2)^{2}-4(5)(-10)$

$=4+200=204>0$

Therefore, the equation $5 x^{2}-2 x-10=0$ has two distinct real roots.

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

$=\frac{-(-2) \pm \sqrt{204}}{2 \times 5}=\frac{2 \pm 2 \sqrt{51}}{10}$

$=\frac{1 \pm \sqrt{51}}{5}=\frac{1+\sqrt{51}}{5}, \frac{1-\sqrt{51}}{5}$

(iv) Given equation is $\frac{1}{2 x-3}+\frac{1}{x-5}=1, x \neq \frac{3}{2}, 5$

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

$\Rightarrow \quad \frac{3 x-8}{2 x^{2}-5 x-10 x+25}=1$

$\Rightarrow \quad \frac{3 x-8}{2 x^{2}-15 x+25}=1$

$\Rightarrow \quad 3 x-8=2 x^{2}-15 x+25$

$\Rightarrow \quad 2 x^{2}-15 x-3 x+25+8=0$

$\Rightarrow \quad 2 x^{2}-18 x+33=0$

On comparing with $a x^{2}+b x+c=0$, we get

$a=2, b=-18$ and $c=33$

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

$=(-18)^{2}-4 \times 2(33)$

$=324-264=60>0$

Therefore the equation $2 x^{2}-18 x+33=0$ has two distinct real roots.

Roots, $x=\frac{-b \pm \sqrt{D}}{2 a}=\frac{-(-18) \pm \sqrt{60}}{2(2)}$

$=\frac{18 \pm 2 \sqrt{15}}{4}=\frac{9 \pm \sqrt{15}}{2}$

$=\frac{9+\sqrt{15}}{2}, \frac{9-\sqrt{15}}{2}$

(v) Given equation is $x^{2}+5 \sqrt{5} x-70=0$

On comparing with $a x^{2}+b x+c=0$, we get

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

$\therefore \quad$ Discriminant, $D=b^{2}-4 a c=(5 \sqrt{5})^{2}-4(1)(-70)$

$=125+280=405>0$

Therefore, the equation $x^{2}+5 \sqrt{5} x-70=0$ has two distinct real roots.

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

$=\frac{-5 \sqrt{5} \pm \sqrt{405}}{2(1)}=\frac{-5 \sqrt{5} \pm 9 \sqrt{5}}{2}$

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

$=\frac{4 \sqrt{5}}{2},-\frac{14 \sqrt{5}}{2}=2 \sqrt{5},-7 \sqrt{5}$

 

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