How do you solve #x^2 - 6x + 9 = 25#?

4 Answers

Answer:

#x=-2, 8#

Explanation:

Given quadratic equation:

#x^2-6x+9=25#

#x^2-6x-16=0#

#x^2-8x+2x-16=0#

#x(x-8)+2(x-8)=0#

#(x-8)(x+2)=0#

#x-8=0, x+2=0#

#x=8, x=-2#

#x=-2, 8#

Jul 9, 2018

Answer:

#x=-2" or "x=8#

Explanation:

#"subtract 25 from both sides"#

#x^2-6x-16=0larrcolor(blue)"in standard form"#

#"the factors of "-16" which sum to "-6#
#"are "-8" and "+2#

#(x-8)(x+2)=0#

#"equate each factor to zero and solve for "x#

#x+2=0rArrx=-2#

#x-8=0rArrx=8#

Jul 16, 2018

Answer:

#x=8# and #x=-2#

Explanation:

Since we have a quadratic, let's set it equal to zero to find its zeroes. This can be done by subtracting #25# from both sides.

We now have

#x^2-6x-16=0#

To factor this, let's do a little thought experiment:

What two numbers sum up to #-6# and have a product of #-16#? After some trial and error, we arrive at #-8# and #2#.

This means we can factor this as

#(x-8)(x+2)=0#

Setting both factors equal to zero, we get

#x=8# and #x=-2#

Hope this helps!

Jul 16, 2018

Answer:

#x = 8" "# or #" "x = -2#

Explanation:

Given:

#x^2-6x+9=25#

Note that both the left hand side and the right hand side are perfect squares, namely:

#(x-3)^2 = 5^2#

Hence:

#x-3=+-5#

So:

#x = 3+-5#

That is:

#x = 8" "# or #" "x = -2#