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

Question

Kennedy Adams · Accepted Answer

#x=-9 +-sqrt91#

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.

Eva Adamson · Answer

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

#ax^2 + bx + c#

A needs to be 1 (which it is) in order to finish the square, and:

#c=(b/2)^2#

#x^2+18x=10#

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

Observe that in order to maintain the value, we must add c to both sides of the equation.

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

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

Now consider the left side:

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

#(x+9)^2 = 91#

We now resolve:

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

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

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

Eva Adamson · Answer

undefined To solve the equation $x^2 + 18x = 10$ by completing the square, follow these steps:

1. Move the constant term to the other side of the equation:
$$x^2 + 18x - 10 = 0$$

2. Add and subtract the square of half the coefficient of $x$ (the middle term coefficient):
$$x^2 + 18x + 9^2 - 9^2 - 10 = 0$$

3. Rewrite the expression as a perfect square trinomial and combine like terms:
$$(x + 9)^2 - 91 = 0$$

4. Add 91 to both sides of the equation:
$$(x + 9)^2 = 91$$

5. Take the square root of both sides:
$$x + 9 = \pm \sqrt{91}$$

6. Subtract 9 from both sides to isolate $x$:
$$x = -9 \pm \sqrt{91}$$

So the solutions to the equation $x^2 + 18x = 10$ by completing the square are $x = -9 + \sqrt{91}$ and $x = -9 - \sqrt{91}$.