How do you solve #5q^2 + 18q = 8#?

1 Answer
Apr 16, 2016

Answer:

#q = 4, 2/5#

Explanation:

We can use factor by grouping to solve this problem.

But first we need to make sure that the equation is in standard form: #ax^2 + bx + c = 0#.

Therefore, subtract 8 from both sides of the equal sign.
#5q^2 + 18q - 8 = 8 - 8#
#5q^2 + 18q - 8 =0#

We need to multiply the a and the c terms and see what factors of that product will add to 18.

#5(-8) = -40#
Factors of -40 that can add to 18 is -2 and 20.
Check: #-2 times 20 = -40# and #-2 + 20 = 18#.

So we can expand the quadratic to
#5q^2 - 2q + 20q - 8 = 0#

Take the GCF of the first two terms (#5q^2# and -2q) which is #q#. Take the GCF of the second two terms (20q and -8), which is 4.

#q(5q - 2)# and #4(5q - 2)#
Notice how the terms in the parenthesis are the same. Those terms inside of the parenthesis will be one binomial of the factored form of the quadratic.

The outside terms of the parentheses, 4 and q, will be combined into one parenthesis: (#q + 4#)

The factored form is #(q + 4)(5q - 2)#

Since it said to solve it, we need to solve for the roots or what value of q will make the expressions in the parenthesis equal to 0.
#q + 4 = 0#
#q = 4#

#5q - 2 = 0#
#q = 2/5#

The roots of this quadratic are 4 and #2/5#.