Question

plzz solve this fast
factorise x^2+3x+9​

Answer 1

Answer:

$\large \text{ x=\dfrac{-3+3i\sqrt{3} }{2} \ or \ x=\dfrac{-3-3i\sqrt{3}}{2} }$

Step-by-step explanation:

Given :

$\large p(x)= x^2+3x+9$

We have to factorise the p ( x ).

By Sridharacharya method we have

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

Comparing from p ( x ) we have

b =  3

a = 1

c = 9

Now putting values in formula

$\large \text{ x=\dfrac{-3\pm\sqrt{3^2-4\times1\times9} }{2\times1} }\\\\\\\large \text{ x=\dfrac{-3\pm\sqrt{9-36} }{2} }\\\\\\\large \text{ x=\dfrac{-3\pm\sqrt{-27} }{2} }\\\\\\\large \text{ x=\dfrac{-3\pm3\sqrt{-3} }{2} }\\\\\\\large \text{ x=\dfrac{-3+3\sqrt{-3} }{2} \ or \ x=\dfrac{-3-3\sqrt{-3}}{2} }$

We know that $i=\sqrt{-1}$

Take out value of i  we get

$\large \text{ x=\dfrac{-3+3i\sqrt{3} }{2} \ or \ x=\dfrac{-3-3i\sqrt{3}}{2} }$

Thus we get answer.