How do you solve using the completing the square method #8x^2 – 80x = –24#?

Question

Ethan Adkins · Accepted Answer

#x=5+sqrt22# or #x=5-sqrt22# In #8x^2-80x=-24#, we have #8# as common factor. Hence dividing by #8#, we get #x^2-10x=-3#. We can make the Left Hand Side of #x^2-10x#, a complete square by comparing it with #(x-a)^2=x^2-2ax+a^2# i.e. by adding square of half the coefficient of #x#. As coefficient of #x# is #-10#, we need to add #(-10/2)^2=25#, to each side and then we have #x^2-10x+25=25-3=22# or #(x-5)^2=(sqrt22)^2# Hence either #x-5=sqrt22# or #x-5=-sqrt22# i.e. #x=5+sqrt22# or #x=5-sqrt22#

Ethan Adkins · Answer

undefined To solve the quadratic equation $8x^2 - 80x = -24$ using the completing the square method:

1. Move the constant term to the right side of the equation:
   $8x^2 - 80x + 24 = 0$

2. Divide the entire equation by the leading coefficient, if it's not 1, to simplify:
   $x^2 - 10x + 3 = 0$

3. To complete the square, take half of the coefficient of $x$ (which is $-10/2 = -5$), square it, and add it to both sides of the equation:
   $x^2 - 10x + (-5)^2 = 3 + (-5)^2$
   $x^2 - 10x + 25 = 3 + 25$

4. Simplify both sides:
   $x^2 - 10x + 25 = 3 + 25$
   $x^2 - 10x + 25 = 28$

5. Rewrite the left side as a perfect square trinomial:
   $(x - 5)^2 = 28$

6. Take the square root of both sides:
   $x - 5 = \pm \sqrt{28}$

7. Simplify the square root of 28:
   $\sqrt{28} = \sqrt{4 \times 7} = 2\sqrt{7}$

8. Solve for $x$:
   $x = 5 \pm 2\sqrt{7}$

So, the solutions to the equation $8x^2 - 80x = -24$ using the completing the square method are $x = 5 + 2\sqrt{7}$ and $x = 5 - 2\sqrt{7}$.