If #x^2 +17 = 81#, what is the value of #x#?

Question

1281 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-09T21:56:22

The highest power is 2.

Because x has a maximum power of 2, there will be 2 answers. if it were 3, there could be up to 3 answers, etc.

While solving this problem, you gather all terms on one side so the equation looks like

#x^2-81+17=0#

#x^2-64=0#

then factor

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

then, because any number times 0 equals 0, we know that one of the two parts in the parenthesis must be equal to 0, so

#x-8=0# or #x+8=0#

and so the two answers #x=8,-8 # come to be.

Answer 2 · 2018-02-09T21:56:49

We start by putting all terms to one side of the equal sign:

#x^2 - 81 + 17 = 0#

#x^2 - 64= 0#

Note that #(a- b)(a + b) = a^2 - b^2#. This is called the *difference of squares * identity. Therefore, we can rewrite the equation as

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

We can now see that if #x= +8# or #x= -8#, the equation will hold true.

We can also confirm graphically. If we trace the parabola #y_1 = x^2 - 64#, the x-intercepts will be the solution to #0 = x^2 -64#. Let's check:

enter image source here

Therefore, our answer is correct.

Hopefully this helps!