How to solve completing the square? 2x^2-8x-15=0

2 Answers
Ch
Feb 14, 2018

#x=± sqrt(11.5)+2#

Explanation:

#2x^2-8x-15=0#

Completing square method:

  • Separate variable terms from constant term, rearrange the equation:

#2x^2-8x=15#

  • Make sure the coefficient of #x^2# is always 1.
    Divide the equation by 2:

#x^2-4x=7.5#

  • Add 4 to left, completing square.

#x^2-4x+4=11.5#

  • Factor the expression on the left

#(x-2)^2=11.5#

  • Take the square root

#sqrt((x-2)^2)=± sqrt(11.5)#

#x-2=± sqrt11.5#

#x=± sqrt(11.5)+2# or #x=± sqrt(23/2)+2#

Sherri
Feb 14, 2018

Answer: #2+- sqrt(11.5)#

Explanation:

#2x^2-8x-15=0#

As we are completing the square of more than one #x^2#, it is best to move the constant (15) to the other side. It's sign therefore, changes - (15 not -15).

#2x^2-8x=15#

Now we divide through by two, to obtain a single #x^2#

#x^2-4x=7.5#

To complete the square, the general steps are to take half the coefficient of x. In this case, the coefficient is 4 therefore half is two. We form brackets, leaving:

#(x-2)^2#

But, if we multiplied this out we would end up with #x^2-4x+4#
We don't want this 'extra' 4, so to complete the square, we must SUBTRACT 4, leaving;

#(x-2)^2-4=7.5#

Now we solve like a standard linear equation;
#(x-2)^2=7.5+4#
#(x-2)^2=11.5#
#x-2=+-sqrt(11.5)#
#x=2+-sqrt(11.5)#

Remember: when you move across the equals sign, you carry out the opposite operation
i.e square, square root
add, subtract
multiply, divide.

Also, when you square root a number you get both a positive AND negative number.

Hope this helps!

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