Question

1. $x^2 + 18x + 81 = 25$

a. 14, 4 c. 14, -14
b. -4, -14 d. -4, 4

use the quadratic formula to solve the equation.

1. $-x^2 + 6x - 5 = 0$

a. -5, - c. -5, 11
b. 1, 5 d. 2, 10

1. $-2x^2 - 5x + 5 = 0$

a. $\displaystyle -\frac{5}{2} \pm \frac{\sqrt{65}}{2}$ c. $\displaystyle -\frac{4}{5} \pm \frac{\sqrt{130}}{4}$
b. $\displaystyle -\frac{5}{4} \pm \frac{\sqrt{32}}{2}$ d. $\displaystyle -\frac{5}{4} \pm \frac{\sqrt{65}}{4}$

1. $-4x^2 + x = -4$

a. $\displaystyle 8 \pm \frac{\sqrt{65}}{8}$ c. $\displaystyle 8 \pm \frac{\sqrt{130}}{8}$
b. $\displaystyle \frac{1}{4} \pm \frac{\sqrt{65}}{4}$ d. $\displaystyle \frac{1}{8} \pm \frac{\sqrt{65}}{8}$

Explanation:

Problem 32

Step1: Rewrite the equation

The equation \(x^{2}+18x + 81=25\) can be rewritten as \(x^{2}+18x+81 - 25=0\), so \(x^{2}+18x + 56 = 0\). Alternatively, we can notice that \(x^{2}+18x + 81=(x + 9)^{2}\), so \((x + 9)^{2}=25\).

Step2: Solve for \(x\)

Take square roots on both sides: \(x+9=\pm5\).

• When \(x + 9 = 5\), \(x=5 - 9=-4\).
• When \(x + 9=-5\), \(x=-5 - 9=-14\).

Step1: Rewrite the equation

The equation \(-x^{2}+6x - 5 = 0\) can be multiplied by \(- 1\) to get \(x^{2}-6x + 5 = 0\).

Step2: Use quadratic formula or factor

We can factor the quadratic: \(x^{2}-6x + 5=(x - 1)(x - 5)=0\). So \(x-1 = 0\) or \(x - 5=0\), which gives \(x = 1\) or \(x = 5\).

Step1: Identify \(a,b,c\)

For the equation \(-2x^{2}-5x + 5 = 0\), we can rewrite it as \(2x^{2}+5x - 5 = 0\) (multiplying by \(- 1\)). Here, \(a = 2\), \(b = 5\), \(c=-5\).

Step2: Apply quadratic formula

The quadratic formula is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).
Calculate the discriminant \(\Delta=b^{2}-4ac=(5)^{2}-4\times2\times(-5)=25 + 40 = 65\).
Then \(x=\frac{-5\pm\sqrt{65}}{2\times2}=\frac{-5\pm\sqrt{65}}{4}\)? Wait, no, wait. Wait, if we use the original equation \(-2x^{2}-5x + 5 = 0\), then \(a=-2\), \(b=-5\), \(c = 5\).
Quadratic formula: \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}=\frac{5\pm\sqrt{(-5)^{2}-4\times(-2)\times5}}{2\times(-2)}=\frac{5\pm\sqrt{25 + 40}}{-4}=\frac{5\pm\sqrt{65}}{-4}=-\frac{5}{4}\pm\frac{\sqrt{65}}{4}\)? Wait, no, let's recalculate.
Wait, the original equation is \(-2x^{2}-5x + 5 = 0\), so \(a=-2\), \(b=-5\), \(c = 5\).
Discriminant \(\Delta=b^{2}-4ac=(-5)^{2}-4\times(-2)\times5=25 + 40 = 65\).
Then \(x=\frac{-(-5)\pm\sqrt{65}}{2\times(-2)}=\frac{5\pm\sqrt{65}}{-4}=-\frac{5}{4}\pm\frac{\sqrt{65}}{4}\)? Wait, no, the option a is \(-\frac{5}{2}\pm\frac{\sqrt{65}}{2}\). Wait, maybe I made a mistake in sign. Let's use the original equation \(-2x^{2}-5x + 5 = 0\), so \(2x^{2}+5x - 5 = 0\) (multiply by - 1). Then \(a = 2\), \(b = 5\), \(c=-5\). Then \(x=\frac{-5\pm\sqrt{25+40}}{4}=\frac{-5\pm\sqrt{65}}{4}\)? No, wait \(2a = 4\)? No, \(a = 2\), so \(2a=4\)? Wait, no, quadratic formula is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). For \(2x^{2}+5x - 5 = 0\), \(a = 2\), \(b = 5\), \(c=-5\). So \(x=\frac{-5\pm\sqrt{25-4\times2\times(-5)}}{2\times2}=\frac{-5\pm\sqrt{25 + 40}}{4}=\frac{-5\pm\sqrt{65}}{4}\). But the option a is \(-\frac{5}{2}\pm\frac{\sqrt{65}}{2}\). Wait, maybe the original equation is \(-2x^{2}-5x + 5 = 0\) and we use \(a=-2\), \(b=-5\), \(c = 5\). Then \(x=\frac{5\pm\sqrt{25+40}}{-4}=\frac{5\pm\sqrt{65}}{-4}=-\frac{5}{4}\pm\frac{\sqrt{65}}{4}\). But the option d is \(-\frac{5}{4}\pm\frac{\sqrt{65}}{4}\). Wait, maybe I messed up. Wait, let's check the original equation again. The equation is \(-2x^{2}-5x + 5 = 0\). So \(a=-2\), \(b=-5\), \(c = 5\). Then discriminant \(D=b^{2}-4ac=(-5)^{2}-4\times(-2)\times5=25 + 40 = 65\). Then \(x=\frac{-b\pm\sqrt{D}}{2a}=\frac{5\pm\sqrt{65}}{2\times(-2)}=\frac{5\pm\sqrt{65}}{-4}=-\frac{5}{4}\pm\frac{\sqrt{65}}{4}\), which is option d. Wait, but the option a is \(-\frac{5}{2}\pm\frac{\sqrt{65}}{2}\). Maybe there is a miscalculation. Wait, if we use the equation as \(-2x^{2}-5x + 5 = 0\), then \(2x^{2}+5x - 5 = 0\), \(a = 2\), \(b = 5\), \(c=-5\). Then \(x=\frac{-5\pm\sqrt{25 + 40}}{4}=\frac{-5\pm\sqrt{65}}{4}\), which is \(-\frac{5}{4}\pm\frac{\sqrt{65}}{4}\), so option d.

Answer:

b. \(-4,-14\)

Problem 33