## Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole

# Chapter 1 - Section 1.5 - Equations - 1.5 Exercises: 61

#### Answer

$x=-2\pm\sqrt{\dfrac{7}{2}}$

#### Work Step by Step

$2x^{2}+8x+1=0$ Let's take the independent term to the right side of the equation: $2x^{2}+8x=-1$ Take out common factor $2$ on the left side of the equation: $2(x^{2}+4x)=-1$ Let's complete the square. Remember that, in order to do that, we need to add $(\dfrac{b}{2})^{2}$ to both sides of the equation. $b$ is always the coefficient of the first degree term. In this equation $b=4$ Please note that, since the expression whose square needs to be completed is multiplied by $2$, we will add $2(\dfrac{b}{2})^{2}$ to the right side of the equation. $2[x^{2}+4x+(\dfrac{4}{2})^{2}]=-1+2(\dfrac{4}{2})^{2}$ $2(x^{2}+4x+4)=-1+8$ $2(x^{2}+4x+4)=7$ $x^{2}+4x+4=\dfrac{7}{2}$ We have a perfect square trinomial on the left side of the expression. We factor it and the equation becomes: $(x+2)^{2}=\dfrac{7}{2}$ Take the square root of both sides of the equation: $\sqrt{(x+2)^{2}}=\sqrt{\dfrac{7}{2}}$ $x+2=\pm\sqrt{\dfrac{7}{2}}$ Solve for $x$: $x=-2\pm\sqrt{\dfrac{7}{2}}$

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