# How do you solve 15a^2=60a?

Dec 18, 2016

$a = 0 \text{ or } a = 4$

#### Explanation:

This is a quadratic equation, set it equal to 0.

$15 {a}^{2} - 60 a = 0 \text{ } \leftarrow$ divide by 15 on both sides.

${a}^{2} - 4 a = 0 \text{ }$factor out the HCF which is $a$

$a \left(a - 4\right) = 0$

Set each factor equal to 0

$a = 0 \text{ or } a - 4 = 0$

$a = 0 \text{ or } a = 4$

Dec 18, 2016

$\textcolor{red}{\text{This a trap: You think that there is only 1 value for } a}$

Very wrong! $\text{ "a=0" or } a = 4$

#### Explanation:

$\textcolor{b l u e}{\text{The trap}}$

divide both sides by $a$

$15 a = 60$

Divide both sides by 15

$a = \frac{60}{15} = \frac{12}{3} = 4$

$a = 4$
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$\textcolor{b l u e}{\text{The better way}}$

Divide both sides by 15

${a}^{2} = 4 a$

$\implies {a}^{2} - 4 a = 0$

$a \left(a - 4\right) = 0$

$a = 0 \text{ or } a = 4$