Question

what are the roots of the equation x^2+6x-9=0

Answer 1

Answer:

The roots of the quadratic equation are

$x=-3+3\sqrt{2}$

$x=-3-3\sqrt{2}$

Step-by-step explanation:

we have

$x^{2} +6x-9=0$

we know that

The formula to solve a quadratic equation of the form $ax^{2} +bx+c=0$ is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$x^{2} +6x-9=0$

so

$a=1\\b=6\\c=-9$

substitute in the formula

$x=\frac{-6(+/-)\sqrt{6^{2}-4(1)(-9)}} {2(1)}$

$x=\frac{-6(+/-)\sqrt{72}} {2}$

$x=\frac{-6(+/-)6\sqrt{2}} {2}$

$x=-3(+/-)3\sqrt{2}$

$x=-3(+)3\sqrt{2}$

$x=-3(-)3\sqrt{2}$

Answer 2

Given equation is :

$x^{2}+6x-9=0$

We will solve this quadratic equation, using the formula:

$x1,2=\frac{-b+\sqrt{b^{2}-4ac}}{2a}$ and

$x1,2=\frac{-b-\sqrt{b^{2}-4ac}}{2a}$

Now, putting a=1 , b=6 and c =-9 we get

$x=\frac{-6+\sqrt{6^{2}-4*1(-9)}}{2*1}$ and

$x=\frac{-6-\sqrt{6^{2}-4*1(-9)}}{2*1}$

Final solutions are:  $x=3(\sqrt{2}-1)$ and

$x=-3(\sqrt{2}+1)$