If a x^2 + bx + c = 0, then x = - Vidyadip

MCQ

If $a{x^2} + bx + c = 0$, then $x =$

A
$\frac{{b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
B
$\frac{{ - b \pm \sqrt {{b^2} - ac} }}{{2a}}$
✓
$\frac{{2c}}{{ - b \pm \sqrt {{b^2} - 4ac} }}$
D
None of these

✓

Answer

Correct option: C.

$\frac{{2c}}{{ - b \pm \sqrt {{b^2} - 4ac} }}$

c
(c) $x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Since $\frac{{2c}}{{ - b + \sqrt {{b^2} - 4ac} }}\,\,.\,\,\frac{{ - b - \sqrt {{b^2} - 4ac} }}{{ - b - \sqrt {{b^2} - 4ac} }}$

=$\frac{{2c\,( - b - \sqrt {{b^2} - 4ac} )}}{{4ac}} = \frac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}$
Similarly $\frac{{2c}}{{ - b - \sqrt {{b^2} - 4ac} }} \times \frac{{ - b + \sqrt {{b^2} - 4ac} }}{{ - b + \sqrt {{b^2} - 4ac} }}$

=$\frac{{2c\,( - b + \sqrt {{b^2} - 4ac} )}}{{4ac}} = \,\frac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}}$

Aliter : On rationalising the given equation

$x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$, we get option $(c)$ correct.

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