Question

5 x square + 4 root 10 x + 8 factorisation​

Answer 1

Question :-

$\large \rm{factorise \: it : } \\ \to \rm 5 {x}^{2} + 4 \sqrt{10} x + 8 = 0$

Answer :-

$\to \boxed{ \rm x = \frac{ - 4}{ \sqrt{10} } }$

Used Formula :-

→ Here we use Shridharacharya formula when a quadratic is given by ax² + bx + c = 0 ,so

$\to \boxed{ \rm \: x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a} }$

Solution :-

We have ,

$\to \: \rm 5 {x}^{2} + 4 \sqrt{10} x + 8 = 0 \: \\ \sf here \\ \to \rm a = 5 \:, \: b = 4 \sqrt{10} \: and \: c = 8 \\ \\ \sf \red{ now \: apply \: the \: given \: formula \: } \\ \\ \to \rm x = \frac{ - 4 \sqrt{10} \pm \sqrt{ {(4 \sqrt{10}) }^{2} - 4 \times 5 \times 8} }{2 \times 5} \\ \\ \to \rm \: x = \frac{ - 4 \sqrt{10} \pm \sqrt{16 \times 10 - 160} }{10} \\ \\ \to \rm \: x = \frac{ - 4 \sqrt{10} \pm \sqrt{160 - 160} }{10} \\ \\ \to \rm x = \frac{ - 4 \sqrt{10} }{ \sqrt{10} \sqrt{10} } \\ \\ \to \boxed{\rm x = \frac{ - 4}{ \sqrt{10} } } \\ \\ \boxed{\bf \underline{verification} \: } : \\ \\ \rm put \rm \: x = \frac{ - 4}{ \sqrt{10} } \\ \\ \to \: 5 { \bigg( \frac{ - 4}{ \sqrt{10} } \bigg)}^{2} + 4 \sqrt{10} \bigg ( \frac{ - 4}{ \sqrt{10} } \bigg) + 8 = 0 \\ \\ \to \: 5 \times \frac{16}{10} - 4 \times 4 + 8 = 0 \\ \\ \to \: 8 - 16 + 8 = 0 \\ \\ \to \: 0 = 0 \: \\ \bf \red{ hence \: verified \: }$