How do you solve #x^2+18x=10# by completing the square?

2 Answers

Answer:

#x=-9 +-sqrt91#

Explanation:

For completing the square, you take half the coefficient of the x value, square it, and add it. So in this case, 81 I suppose?
I apologize, I don't quite remember how to do this. However there is already a 10...

Oh! I got it!

So add 91 and then you would get:

#x^2+18x+81=91#

Then you factor:

#(x+9)^2=91#

Square root:

#x+9=+- sqrt91#

Solve by subtracting 9:

#x=-9+-sqrt91#

And that should be your answer.

Jun 7, 2018

Answer:

#x= -9+-sqrt(91)#

Explanation:

#ax^2+bx +c#

To complete the square a must be 1 (which it is) and:

#c=(b/2)^2#

#x^2+18x=10#

#x^2+18x + c =10+c#

notice we have to add c to both sides of the equation so we don't alter the value.

#c=(18/2)^2= 81#

#x^2+18x + 81 =10+81#

now factor the left side:

#(x+9)(x+9) = 91#

#(x+9)^2= 91#

now we solve:

#sqrt((x+9)^2)= +-sqrt(91)#

#x+9= +-sqrt(91)#

#x= -9+-sqrt(91)#