Question

Find roots of quadratic equation x² + 2x- 7 = 0 (use formula method).​

Answer 1

$\huge\underline{\bf \orange {Solution :}}$

By using quadratic formula

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

Here

• a = 1
• b = 2
• c = - 7

Substitute values in formula

$\implies\sf x = \dfrac{ - 2\pm \sqrt{ {2}^{2} - 4 \times 1 \times ( - 7)} }{2 \times 1} \\ \\ \implies\sf x = \dfrac{ - 2\pm \sqrt{4 + 28} }{2} \\ \\\implies \sf x = \dfrac{ - 2\pm \sqrt{32} }{2} \\ \\ \bf Taking \: +ve \: \: sign \\ \\ \implies \boxed{\sf x = \dfrac{ - 2 + \sqrt{32} }{2}} \\ \\ \bf Taking \: - ve \: \: sign \\ \\ \implies \boxed{\sf x = \dfrac{ - 2 - \sqrt{32} }{2}}$

Answer 2

${\huge{\bf{\green{\mathfrak{\underline{\underline{Solution}}}}}}}$

By using quadratic formula,

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

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⟹ x= −b± root b^2 −4ac/2a

Here,

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀•a = 1

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀•b = 2

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀•c = - 7

Substitute values in formula

\begin{gathered}\implies\sf x = \dfrac{ - 2\pm \sqrt{ {2}^{2} - 4 \times 1 \times ( - 7)} }{2 \times 1} \\ \\ \implies\sf x = \dfrac{ - 2\pm \sqrt{4 + 28} }{2} \\ \\\implies \sf x = \dfrac{ - 2\pm \sqrt{32} }{2} \\ \\ \bf Taking \: +ve \: \: sign \\ \\ \implies \boxed{\sf x = \dfrac{ - 2 + \sqrt{32} }{2}} \\ \\ \bf Taking \: - ve \: \: sign \\ \\ \implies \boxed{\sf x = \dfrac{ - 2 - \sqrt{32} }{2}}\end{gathered}

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⟹x= −2± root 2^2−4×1×(−7)/2×1

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⟹x= 2−2± 4+28

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⟹x= 2−2± 32

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀Taking +ve sign

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⟹ x= −2+ root 32 /2

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀Taking −ve sign

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⟹ x= −2− root 32 /2