Question

4(x−3)^2−252=0
ye know what im sayin, I need it urgennnnt

Answer 1

`x_1=-4.9373\\ \\x_2=10.9373`

Explanation:

I'm assuming you need the solution of this equation, let's solve it!

1. Write the expression.

`4(x-3)^2-252=0`

2. Solve the square parenthesis.

We are using the following property:

`(a-b)^2=a^2-2ab+b^2`

Let x be "a".

Let 3 be "b"

`(x-3)^2=\\ \\x^2-2x(3)+3^2=\\ \\x^2-6x+9`

3. Take the result of the expansion and substitute it inside the parenthesis in the original expression.

`4(x^2-6x+9)-252=0`

4. Make the associative multiplifation with 4.

`(4)(x^2)+(4)(-6x)+(4)(9)-252=0\\ \\(4x^2)+(-24x)+(36)-252=0\\ \\4x^2-24x-216=0`

5.Simplify the expression.

`(4x^2)/(4) -(24x)/(4) -(216)/(4) =(0)/(4) \\ \\x^2-6x-54=0`

6. Find the roots.

Let's do it by applying the formula for solving quadratic equation. This is the formula:

`x=(-b(+-)√(b^2-4ac) )/(2a)`

If you take a closer look, the equation has a ± symbol on the upper part to the right of -b, this means that either sum or substraction can be done to solve this equation. This means that we are going to get 2 solutions, always. Hence:

`x_1=(-b-√(b^2-4ac) )/(2a)\\ \\x_2=(-b+√(b^2-4ac) )/(2a)`

Tha values of a, b, and c are the following:

Since x² is being multiplied by 1, a= 1;

Since -6x is being multiplied by -6, b= -6;

Since -54 is a constant, c= .

Substitute in the 2 equations and find the roots (could be done with a calculator).

`x_1=(-(-6)-√((-6)^2-4(1)(-54)) )/(2(1))=-4.9373\\ \\x_2=x_1=(-(-6)+√((-6)^2-4(1)(-54)) )/(2(1))=10.9373`

7. Express your results.

`x_1=-4.9373\\ \\x_2=10.9373`.