Question

What is x if `-4(x+2)^2=-20`?

Answer 1

`" "`
Approximate values of `color(red)(x)` are:

`color(blue)(x~~0.236`

`color(blue)(x~~-4.236`

Explanation:

`" "`
How do we find the value of `color(red)(x` using

`color(red)([-4(x+2)^2]=(-20)`

`rArr (-4)(x^2+4x+4)=(-20)`

Divide both sides of the equation by `color(red)((-4)`

`rArr [(-4)(x^2+4x+4)]/((-4))=((-20))/((-4)`

`rArr [cancel (-4)(x^2+4x+4)]/(cancel ((-4)))=((-20))/((-4)`

`rArr x^2+4x+4=5`

Subtract `color(red)(5)` from both sides of the equation.

`rArr x^2+4x+4-5=5-5`

`rArr x^2+4x+4-5=cancel 5-cancel 5`

`rArr x^2+4x-1=0` ...Eqn.1

We now have a quadratic equation.

Use the quadratic formula to find the values of `color(red)(x`

Quadratic formula:

`color(blue)(x=[-b+-sqrt(b^2-4ac)]/(2a)`

Using Eqn.1, we get

`a=1; b=4 and c= (-1)`

Substitute the values in the quadratic formula above:

`color(blue)(x=[-4+-sqrt(4^2-4*1*(-1))]/(2*1)`

`x=[-4+- sqrt(16+4)]/2`

`x=[-4+-sqrt(20)]/2`

`x=[-4+-sqrt(4*5)]/2`

`x=[-4+- sqrt(4)*sqrt(5)]/2`

`x=[-4+-2*sqrt(5)]/2`

`x=(-4/2)+-(2*sqrt(5))/2`

`x=(-2)+-(cancel 2*sqrt(5))/cancel 2`

`x=-2+-sqrt(5)`

`x=-2-sqrt(5)`, `x=-2+sqrt(5)`

Using a spreadsheet software or a calculator, we get

Hence, approximate values of `color(red)(x)` are:

`color(blue)(x~~0.236`

`color(blue)(x~~-4.236`

Hope it helps.

Answer 2

`x=-2+sqrt5` and `-2-sqrt5`

Explanation:

Let's start by dividing both sides by `-4`. We now have

`(x+2)^2=5`

Let's take the square root of both sides to get

`x+2=sqrt5` and `x+2=-sqrt5`

Subtracting `2` from both sides of both equations, we get

`x=-2+sqrt5` and `-2-sqrt5`

Hope this helps!

Answer 3

`x = -2 +- sqrt5`

Explanation:

`-4(x+2)^2 = -20`

First, divide both sides by `color(blue)(-4)`:
`(-4(x+2)^2)/color(blue)(-4) = (-20)/color(blue)(-4)`

`(x+2)^2 = 5`

Expand/simplify the left hand side:
`x^2 + 4x + 4 = 5`

Subtract `color(blue)5` from both sides:
`x^2 + 4x + 4 quadcolor(blue)(-quad5) = 5 quadcolor(blue)(-quad5)`

`x^2 + 4x - 1 = 0`

This is now in standard form, `ax^2 + bx + c`.

Use the quadratic formula `(-b +- sqrt(b^2 - 4ac))/(2a)` to find the value of `x`:
`x = (-4 +- sqrt(4^2 - 4(1)(-1)))/(2(1))`

`x = (-4 +- sqrt(16 + 4))/2`

`x = (-4 +- sqrt20)/2`

`x = (-4 +- 2sqrt5)/2`

`x = -2 +- sqrt5`

Hope this helps!