Question

2. 25x2 + 50x + 12 = 0. samikaran ke mul sutra dwara gyat kijiye​

Answer 1

GIVEN :–

• A quadratic equation 25x² + 50x + 12 = 0.

TO FIND :–

• Roots = ?

SOLUTION :–

• If a quadratic equation ax² + bx + c = 0 have two roots , then roots are –

$\\ \longrightarrow \large{ \boxed{ \sf x = \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a} }}$

• Here –

$\\ \sf \: \: \: \: { \huge{.}} \: \: a = 25$

$\\ \sf \: \: \: \: { \huge{.}} \: \: b = 50$

$\\ \sf \: \: \: \: { \huge{.}} \: \: c = 12$

• put the values –

$\\ \implies \sf x = \dfrac{ - (50) \pm \sqrt{ {(50)}^{2} - 4(25)(12)} }{2(25)}$

$\\ \implies \sf x = \dfrac{ - 50 \pm \sqrt{ 2500 - 1200}}{2(25)}$

$\\ \implies \sf x = \dfrac{ - 50 \pm \sqrt{ 1300}}{50}$

$\\ \implies \sf x = \dfrac{ - 50 \pm 10 \sqrt{ 13}}{50}$

$\\ \implies \sf x = \dfrac{- 5\pm \sqrt{ 13}}{5}$

$\\ \implies \sf x = \dfrac{ - 5 + \sqrt{ 13}}{5} \: , \: \frac{ - 5 - \sqrt{13} }{5}$

• Hence the roots are $\: \: \sf \dfrac{ - 5 + \sqrt{ 13}}{5} \:, \: \dfrac{ - 5 - \sqrt{13} }{5}$