Question

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

Answer 1

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(√(2)-1)` and

`x=-3(√(2)+1)`

Answer 2

The roots of the quadratic equation are

`x=-3+3√(2)`

`x=-3-3√(2)`

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(+/-)√(72)} {2}`

`x=\frac{-6(+/-)6√(2)} {2}`

`x=-3(+/-)3√(2)`

`x=-3(+)3√(2)`

`x=-3(-)3√(2)`