Question

The zeroes of the following polynomial will be x2+88x+125

Answer 1

Step-by-step explanation:

${\red{\underline{\underline{\bold{Given:-}}}}}$

• The quadratic polynomial ${x}^{2}+88x+125$

${\blue{\underline{\underline{\bold{To\:Find:-}}}}}$

• The value of x( The zeroes if the polynomial)

${\green{\underline{\underline{\bold{Solution:-}}}}}$

${x}^{2} + 88x + 125$

By using quadratic formula:-

D = {b}^{2} - 4ac

$\frac{-b\pm\sqrt{D}}{2a}$

Comparing with our equation,

where a=1 , b=88, c=125

$\frac{ - 88 \pm \sqrt{ {( - 88)}^{2} - 4 \times 1 \times 125} } {2 \times 1} \\ \\ = \frac{ - 88 \pm \sqrt{7744 - 500} }{2} \\ \\ = \frac{ - 88 \pm \sqrt{7244} }{2} \\ \\ = \frac{ - 88 \pm2 \sqrt{1811} }{2}$

Take 2 as the common term in the numerator:-

$\frac{2( - 44 \pm \sqrt{1811)} }{2} \\ \\ = - 44 \pm \sqrt{1811}$

Therefore, The roots of the equation are

$\boxed{- 44 + \sqrt{1811} \: \: and \: \: \: - 44 - \sqrt{1811}}$