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To solve this equation take the square root of each side.
-2 and 1
Notice that on the left hand side of the given equation we have a perfect square trinomial. To solve a quadratic equation in the form x^2=n, we will take the square root of each side. For any number n≥ 0, if x^2=n then x=±sqrt(n). Keeping this in mind, let's consider the given equation.
a^2+2ab+b^2=(a+b)^2
sqrt(LHS)=sqrt(RHS)
LHS-1=RHS-1
Calculate root
.LHS /2.=.RHS /2.
We can simplify this result into two separate roots.
| c=- 1 ±3/2 | |
|---|---|
| c_1=- 1-3/2 | c_2=- 1+3/2 |
| c_1=- 4/2 | c_2=2/2 |
| c_1=- 2 | c_2=1 |
We found that the solutions to the given equation are - 2 and 1. To check our answer, we will graph the related function y=4c^2+4c-8 using a calculator. Note that in the calculator we will use the variable x instead of c.
We can see that the x-intercepts are - 2 and 1. Therefore, our solutions are correct.