How do you solve #x^2+10x-1=0# by completing the square?

Question

Savannah Anderson · Accepted Answer

#x=+-sqrt26-5#

Isolate all terms involving #x# on one side and move the constant to the other side.

#x^2+10x=1#

Now, the original quadratic, #x^2+10x-1,# is in the form #ax^2+bx+c# where #a=1, b=10, c=-1#.

To complete the square, we first want to add #(b/2)^2# to each side.

#(b/2)^2=(10/2)^2=5^2=25#, so we add #25# to each side.

#x^2+10x+25=1+25#

The left side needs to be factored. Fortunately, since we added #(b/2)^2# to each side, the factored form will simply be #(x+b/2)^2=(x+5)^2#

#(x+5)^2=26#

Now, take the root of both sides, taking into consideration the right answers, both positive and negative.

#sqrt((x+5)^2)=+-sqrt26#

#x+5=+-sqrt26#

Solving for #x# yields

#x=+-sqrt26-5#

Hazel Anglin · Answer

undefined To solve the quadratic equation x^2 + 10x - 1 = 0 by completing the square, follow these steps:

1. Move the constant term to the other side of the equation: x^2 + 10x = 1
2. Take half of the coefficient of x, square it, and add it to both sides of the equation: x^2 + 10x + (10/2)^2 = 1 + (10/2)^2
3. Simplify both sides: x^2 + 10x + 25 = 1 + 25
4. Rewrite the left side as a perfect square: (x + 5)^2 = 26
5. Take the square root of both sides: x + 5 = ±√26
6. Subtract 5 from both sides: x = -5 ±√26