Question

Solve the equation x^2 + 8x - 16 = 0

justaguide · Accepted Answer

The roots of any quadratic equation of the form ax^2 + bx + c = 0 are `x = (-b+- sqrt(b^2-4ac))/(2a)` .

For the given equation x^2 + 8x - 16 = 0, a = 1, b = 8 and c = -16. The roots of the equation are `(-8+-sqrt(64+64))/2`

= `-4 +- 4*sqrt 2`

The roots of the equation x^2 + 8x - 16 = 0 are `-4 +- 4*sqrt 2`

sciencesolve · Answer

You also may use the following approach to evaluate the solutions to quadratic equation, hence, you may complete the square `x^2 + 8x` using the special product `(a+b)^2 = a^2 + 2ab + b^2` , such that:

`a^2 = x^2 => a = x`

`2ab = 8x => 2*x*b = 8*x => b = 4 => b^2 = 16`

You need to add 16 to the left, to complete the square, and you need to add 16 to the right to preserve the equation, such that:

`x^2 + 8x + 16 - 16 = 0 + 16`

`(x + 4)^2 - 16 = 16 => (x + 4)^2 = 32 => x + 4 = +-sqrt(32)`

`x + 4 = +-4sqrt2 => x_(1,2) = -4 +- 4sqrt2`

Hence, evaluating the solutions to the quadratic equation, completing the square, yields two conjugates `x_1 = -4 + 4sqrt2` and `x_2 = -4 - 4sqrt2` .

jlsmath · Answer

You could use completing the square or the quadratic formula, x=(-b +-sqrt(b^2 - 4ac))/2a, to solve this problem.

Using the quadratic formula, note that in the equation, x^2 + 8x - 16 = 0, a=1, b=8 and c= -16.

x= (-8 +-sqrt(8^2 - (4)(1)(-16)))/2(1)

x= (-8 +-sqrt(64 - (-64))/2(1)

x= (-8 +- sqrt(128))/2

x = (-8+- 8 sqrt (2)) /2

Therefore, the roots are -4 +- 4 sqrt (2).

aruv · Answer

Your question is

`x^2+8x-16=0` ,

you want to solve it for x.

`x^2+8x+16=16+16`

`(x+4)^2=32`

`x+4=+-sqrt(32)`

`x=-4+-sqrt(32)`

`x=-4+-4sqrt(2)`

`x=4(-1+-sqrt(2))`

`x=-4(1+-sqrt(2))`