Write the following equation in standard form. Then solve. 4q^2+2q=3q^2-4q+16

1 answer:

3 0

$4q^2+2q=3q^2-4q+16\qquad\text{subtract}\ 3q^2\ \text{from both sides}\\\\q^2+2q=-4q+16\qquad\text{add}\ 4q\ \text{to both sides}\\\\q^2+6q=16\qquad\text{subtract 16 from both sides}\\\\q^2+6q-16=0\to\boxed{\text{standard form}}\\\\q^2+8q-2q-16=0\\\\q(q+8)-2(q+8)=0\\\\(q+8)(q-2)=0\iff q+8=0\ \vee\ q-2=0\\\\q+8=0\qquad\text{subtract 8 from both sides}\\\\\boxed{q=-8}\\\\q-2=0\qquad\text{add 2 to both sides}\\\\\boxed{q=2}\\\\Answer:\ \boxed{q=-8\ or\ q=2}$

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This can be solved by factoring.

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Then, find two the factors of $ac$ whose sum is $b$ .

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Split $b$ into these two factors.

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Next, factor by grouping.

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By the Zero Product Property, set each factor equal to zero.

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