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

Question

asked Jun 6, 2017 114k views

2 Answers

Salman Nazir · Answer 1 · 2017-06-08T09:12:42+0000

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)

Pduey · Answer 2 · 2017-06-11T19:16:16+0000

Answer:

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)