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

Answer»

Answer:

\large \text{$<klux>X</klux>=\dfrac{-3+3i\sqrt{3} }{2} \ or \ x=\dfrac{-3-3i\sqrt{3}}{2}$}

Step-by-step explanation:

Given :

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

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.


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