How do you solve # x^2 + 14x - 15 = 0# by completing the square?

Question

Eva Adamson · Accepted Answer

The solutions are:
#color(green)(x = 1# or # color(green)(x = -15# #x^2 + 14x - 15 = 0# #x^2 + 14x = 15# We add #49 # to both sides in order to write the Left Hand Side as a Perfect Square: 15 + 49 # = #x^2 + 14x + 49 #2*x^2 + 7 * 7^2 = 64# Using #color(blue)((a+b)^2 = a^2 + 2ab + b^2# as our identity, we obtain #(x+7)^2 = 64# Sqrt64# = #x + 7 or #x + 7 = -sqrt64# Green color #color(x = 8-7 = 1# or #color(green)(x = -8 - 7 = -15#)

Hazel Anglin · Answer

undefined To solve $x^2 + 14x - 15 = 0$ by completing the square, follow these steps:

1. Move the constant term to the other side of the equation: $x^2 + 14x = 15$.
2. Take half of the coefficient of the $x$ term, square it, and add it to both sides of the equation: $x^2 + 14x + (\frac{14}{2})^2 = 15 + (\frac{14}{2})^2$.
3. Simplify both sides: $x^2 + 14x + 49 = 15 + 49$.
4. Combine like terms: $x^2 + 14x + 49 = 64$.
5. Rewrite the left side as a perfect square trinomial: $(x + 7)^2 = 64$.
6. Take the square root of both sides: $x + 7 = \pm \sqrt{64}$.
7. Solve for $x$: $x = -7 \pm 8$.

Therefore, the solutions to the equation $x^2 + 14x - 15 = 0$ by completing the square are $x = -7 + 8$ and $x = -7 - 8$, which simplifies to $x = 1$ and $x = -15$.