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Chapter 11: Problem 2

$$x^{2}+8 x=-7$$ Quantity \(\mathbf{A}\) \(x\) Quantity_B 0 a. Quantity A is greater. b. Quantity B is greater. c. The two quantities are equal. d. The relationship cannot be determined from the information given.

Short Answer

Expert verified
b. Quantity B is greater.

Step by step solution

01

Rearrange the equation to standard form

The standard form of a quadratic equation is \( ax^{2} + bx + c = 0 \). We need to rearrange our equation: \( x^{2} + 8x + 7 = 0 \)
02

Apply the quadratic formula

The quadratic formula is \( x = [-b \pm sqrt(b^{2} - 4ac)] / 2a\). The values for \( a \), \( b \), and \( c \) from our equation are 1, 8, and 7 respectively. Plugging these into the formula, we get two possible solutions: \( x_{1} = (-8 + sqrt[(8^{2} - 4*1*7)]) / 2*1 = -1 \) , \( x_{2} = (-8 - sqrt[(8^{2} - 4*1*7)]) / 2*1 = -7 \)
03

Compare the quantities

Now compare the values of \( x \) to Quantity B, which is 0. Both \( x_1 = -1 \) and \( x_2 = -7 \) are smaller than 0 so Quantity B is greater.

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Most popular questions from this chapter

If \(x=3 a\) and \(y=9 b,\) then all of the following are equal to \(2(x+y)\) EXCEPT a. \(3(2 a+6 b)\) b. \(6(a+3 b)\) c. \(24\left(\frac{1}{4} a+\frac{3}{4} b\right)\) d. \(\frac{1}{3}(18 a+54 b)\) e. \(12\left(\frac{1}{2} a+\frac{3}{4} b\right)\)

If \(A=2 x-(y-2 c)\) and \(B=(2 x-y)-2 c,\) then \(A-B=\) a. \(-2 y\) b. \(-4 c\) c. 0 d. \(2 y\) e. \(4 c\)

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