How do you solve $10b^2 = 27b - 18$ by factoring?

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Answer 1 · 2015-08-21T07:03:36

The solutions are:
$color(blue)(b=6/5$

$color(blue)(b=3/2$

$10b^2-27b+18=0$

We can Split the Middle Term of this expression to factorise it and thereby find solutions.

In this technique we need to think of 2 numbers such that:

$N_1*N_2 = a*c = 10*18 = 180$
and
$N_1 +N_2 = b =-27$

After trying out a few numbers we get $N_1 = -12$ and $N_2 =-15$

$(-12)*(-15) = 180$ , and $(-12)+(-15)= -27$

$10b^2-27b+18=10b^2-15b-12b+18$

$10b^2-15b-12b+18=0$

$5b(2b-3) -6(2b-3)=0$

$(5b-6)(2b-3)=0$ is the factorised form of the expression.

Now we equate the factors to zero.

$5b-6=0, color(blue)(b=6/5$

$2b-3=0, color(blue)(b=3/2$

How do you solve 10b^2 = 27b - 1810b^2 = 27b - 18 by factoring?