How do you solve #x^2+x=1# by completing the square?

Question

Abigail Allen · Accepted Answer

#x^2+x=1#

In order for the left side to be a perfect square, it must be the square of some #x+a#.

Since #(x+a)^2= x^2+2ax+a^2#,

we see that we must have #2a=1#, so #a = 1/2#.

This make #a^2=1/4# which I need on the left to make a perfect square. We'll add 1/4 to bot side to get:

#x^2+x+1/4=1+1/4#

After simplifying the right and factoring the left, we obtain:

#(x+1/2)^2=5/4#.

Now use the square root method to solve:

#x+1/2 = +-sqrt(5/4) = +-sqrt5/2#

So #x=-1/2+-sqrt5/2#, which may also be written:

#x = (-1+-sqrt5)/2#

Savannah Anderson · Answer

undefined To solve the equation $x^2 + x = 1$ by completing the square, follow these steps:

1. Move the constant term to the right side of the equation:
$$x^2 + x - 1 = 0$$

2. Add $\left(\frac{1}{2}\right)^2 = \frac{1}{4}$ to both sides to complete the square on the left side:
$$x^2 + x + \frac{1}{4} - 1 = \frac{1}{4}$$

3. Rewrite the left side as a perfect square trinomial and simplify the right side:
$$\left(x + \frac{1}{2}\right)^2 - 1 = \frac{1}{4}$$

4. Add 1 to both sides:
$$\left(x + \frac{1}{2}\right)^2 = \frac{5}{4}$$

5. Take the square root of both sides, considering both the positive and negative square roots:
$$x + \frac{1}{2} = \pm \sqrt{\frac{5}{4}}$$

6. Simplify:
$$x + \frac{1}{2} = \pm \frac{\sqrt{5}}{2}$$

7. Solve for $x$:
$$x = -\frac{1}{2} \pm \frac{\sqrt{5}}{2}$$