How do you solve $a^{2} - 4 a = 32$ $a^2 - 4a = 32$ ?

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Answer 1 · 2018-04-16T10:00:56

$a = 8, a = - 4$ $a=8, a=-4$

Minus $32$ $32$

$\to a^{2} - 4 a - 32$ $-> a^2-4a-32$

List the factors of $32$ $32$ :

$1$ $1$ and $32$ $32$
$2$ $2$ and $16$ $16$
$4$ $4$ and $8$ $8$

As you can see, we can make $- 4$ $-4$ from $8$ $8$ and $4$ $4$

$\to (a - 8) (a + 4)$ $-> (a-8)(a+4)$

Solve:

$a - 8 = 0$ $a-8=0$

$a = 8$ $a=8$

$a + 4 = 0$ $a+4=0$

$a = - 4$ $a=-4$

Therefore $a = 8, a = - 4$ $a=8, a=-4$

We can always check the answer:

$a = 8$ $a=8$

$8^{2} - (4 \times 8) = 64 - 32 = 32$ $8^2-(4xx8)=64-32=32$

$a = - 4$ $a=-4$

${(- 4)}^{2} - (- 4 \times 4) = 16 + 16 = 32$ $(-4)^2-(-4xx4)=16+16=32$

Hence $a = 8, a = - 4$ $a=8, a=-4$ is correct

How do you solve a2−4a=32a^2 - 4a = 32?