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denis-greek [22]
4 years ago
6

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

Mathematics
1 answer:
forsale [732]4 years ago
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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Answer:

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Step-by-step explanation:

Remember that Taylor says that

f(x) = {\displaystyle \sum\limits_{k=0}^{\infty} \frac{f^{(k)}(a) }{k!}(x-a)^k }

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f^{(0)} (-2) = 8(-2)-3(-2)^3 = 8\\f^{(1)} (-2) = 8-3(3)(-2)^2 = -28\\f^{(2)} (-2) = -3(3)2(-2) = -36\\f^{(2)} (-2) = -3(3)2 = -18

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