How do you solve #x^2+x-72=0#?

Question

Nathan Axel · Accepted Answer

Find a pair of factors of #72# which differ by #1#, hence:

#x^2+x-72 = (x+9)(x-8)#

which has zeros #x=8# and #x=-9# In general #(x+a)(x-b) = x^2+(a-b)x-ab# In our case we want #ab=72# and #a-b = 1# It should not take long to find that #a=9# and #b=8# works.

Hazel Anglin · Answer

Complete the square to find zeros #x=8# and #x=-9#

Alternatively, finish the square like this:

Note that #(x+1/2)^2 = x^2+x+1/4#

We also employ the identity of difference of squares:

#a^2-b^2 = (a-b)(a+b)#

with #a = x+1/2# and #b=17/2#

So:

#x^2+x-72#

#= x^2+x+1/4-1/4-72 = (x+1/2)^2-289/4#

#= (x+1/2)^2-(17/2)^2#

#= ((x+1/2)-17/2)((x+1/2)+17/2)#

#= (x-8)(x+9)#

which has zeros #x=8# and #x=-9#

Isaiah Anthony · Answer

undefined To solve the quadratic equation $x^2 + x - 72 = 0$, you can use the quadratic formula:
$$x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$$
where $a = 1$, $b = 1$, and $c = -72$.
Substitute these values into the formula and solve for $x$.

$$x = \frac{{-1 \pm \sqrt{{1^2 - 4(1)(-72)}}}}{{2(1)}}$$

$$x = \frac{{-1 \pm \sqrt{{1 + 288}}}}{{2}}$$

$$x = \frac{{-1 \pm \sqrt{{289}}}}{{2}}$$

$$x = \frac{{-1 \pm 17}}{{2}}$$

So, the solutions are:
$$x_1 = \frac{{-1 + 17}}{{2}} = 8$$
$$x_2 = \frac{{-1 - 17}}{{2}} = -9$$

Therefore, the solutions to the equation $x^2 + x - 72 = 0$ are $x = 8$ and $x = -9$.