How do you solve $x^{2} + x - 72 = 0$ $x^2+x-72=0$ ?

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How do you solve $x^{2} + x - 72 = 0$ $x^2+x-72=0$ ?

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Answer 1 · 2015-12-23T11:32:29

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

$x^{2} + x - 72 = (x + 9) (x - 8)$ $x^2+x-72 = (x+9)(x-8)$

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

Explanation:

In general $(x + a) (x - b) = x^{2} + (a - b) x - a b$ $(x+a)(x-b) = x^2+(a-b)x-ab$

In our case we want $a b = 72$ $ab=72$ and $a - b = 1$ $a-b = 1$

It should not take long to find that $a = 9$ $a=9$ and $b = 8$ $b=8$ works.

Answer 2 · 2015-12-23T11:39:03

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

Explanation:

Alternatively, complete the square as follows:

Note that ${(x + \frac{1}{2})}^{2} = x^{2} + x + \frac{1}{4}$ $(x+1/2)^2 = x^2+x+1/4$

We also use the difference of squares identity:

$a^{2} - b^{2} = (a - b) (a + b)$ $a^2-b^2 = (a-b)(a+b)$

with $a = x + \frac{1}{2}$ $a = x+1/2$ and $b = \frac{17}{2}$ $b=17/2$

So:

$x^{2} + x - 72$ $x^2+x-72$

$= x^{2} + x + \frac{1}{4} - \frac{1}{4} - 72 = {(x + \frac{1}{2})}^{2} - \frac{289}{4}$ $= x^2+x+1/4-1/4-72 = (x+1/2)^2-289/4$

$= {(x + \frac{1}{2})}^{2} - {(\frac{17}{2})}^{2}$ $= (x+1/2)^2-(17/2)^2$

$= ((x + \frac{1}{2}) - \frac{17}{2}) ((x + \frac{1}{2}) + \frac{17}{2})$ $= ((x+1/2)-17/2)((x+1/2)+17/2)$

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

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

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How do you solve $x^{2} + x - 72 = 0$ $x^2+x-72=0$ ?

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