Question

17) Solve x ^ 2 - 7x + 12 = 0 by using quadratic formula.​

Answer 1

Step-by-step explanation:

${x}^{2} - 7x + 12 = 0 \\ \: \large \sf \blue{ (using \: quadratic \: formula)} \\ {x}^{2} - 4x - 3x + 12 = 0 \\ x(x - 4) - 3(x - 4 ) = 0\\ (x - 3)(x - 4) = 0 \\ \\ \large \sf \blue{x = 4 \: or \:x = 3}$

Answer 2

____________________________________________________. ★ ___________. ★

$\sf\small\red{Given:-}$

$\sf\rightarrow \: {x}^{2} - 7x + 12 = 0$

$\sf\small\red{Find:-}$

$\sf\rightarrow \: solving \: the \: quadratic \: equation.$

$\sf\small\red{formula:-}$

$\sf\rightarrow \: {ax}^{2} + bx + c = 0$

$\sf\small\red{Solution:-}$

$\sf \mapsto\: x = \frac{ - b \pm \: \sqrt{ {b}^{2} } - 4ac }{2a}$

We have,

a = 1.

b = -7.

c = 12.

$\sf\mapsto \: x = \frac{ - ( - 7) \pm \: {( - 7)}^{2} - 4 \: .1. \: 12}{2.1}$

$\sf\mapsto \: \frac{7 \pm \: \sqrt{49 - 48} }{2}$

$\sf\mapsto \: \frac{7 \pm \: \sqrt{1} }{2}$

$\sf\mapsto \: \frac{7 \pm \: 1}{2}$

i.e,

$\sf\Rightarrow \: x = 4 \: or \: 3.$