How do you solve the equation: $5 w^{2} - 5 w = 0$ $5w^2-5w=0$ ?

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Answer 1 · 2018-03-14T17:35:50

Remove extra coefficients to reveal two solutions $w = 1$ $w=1$ and $w = 0$ $w=0$

There is a common factor on the left-hand side that is equal to $5 w$ $5w$ . If you pull that common factor you you see:

$5 w (w - 1) = 0$ $5w(w-1)=0$

Since anything multiplied by zero is zero, we can look at the two scenarios where you get 0=0:

$(w - 1) = 0 \Rightarrow w = 1$ $(w-1)=0 rArr color(red)(w=1)$

$5 w = 0 \Rightarrow w = 0$ $5w=0 rArr color(blue)(w=0)$

We can also treat this like a quadratic equation, where:

$a = 5$ $a=5$
$b = - 5$ $b=-5$
$c = 0$ $c=0$

$w = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $w=(-b+-sqrt(b^2-4ac))/(2a)$

$w = \frac{- (- 5) \pm \sqrt{{(- 5)}^{2} - 4 (5 \cdot 0)}}{2 (5)}$ $w=(-(-5)+-sqrt((-5)^2-4(5*0)))/(2(5))$

$w = \frac{5 \pm \sqrt{25}}{10}$ $w=(5+-sqrt(25))/10$

$w = \frac{5 \pm 5}{10} \Rightarrow w = 1, w = 0$ $w=(5+-5)/10 rArr w=1, w=0$

How do you solve the equation: 5w^2-5w=05w2−5w=05w^2-5w=0?