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Standard X

Mathematics

Quadratic Formula

Find whether ...

Question

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

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Solution

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

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

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

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

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

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

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Quadratic Formula

Standard X Mathematics

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Find whether the following equation 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) 12x−3+1x−5=1,x≠32,5 (v) x2+5√5x−70=0

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