precision solve each equation. if exact roots cannot be found, state the consecutive integers between which the roots are located. 4. $x^2 + 8x = 0$ 5 $x^2 - 3x - 18 = 0$ 6. $4x - x^2 + 8 = 0$ 7. $-12 - 5x + 3x^2 = 0$ 8. $x^2 - 6x + 4 = -8$ 9. $9 - x^2 = 12$ 10. $5x^2 + 10x - 4 = -6$ 11. $x^2 - 20 = 2 + x$

Question

Accepted Answer

### Problem 4: $ x^2 + 8x = 0 $
# Explanation:
## Step1: Factor out $ x $
$ x(x + 8) = 0 $
## Step2: Apply zero - product property
If $ ab = 0 $, then either $ a = 0 $ or $ b = 0 $. So $ x = 0 $ or $ x+8 = 0 $
## Step3: Solve for $ x $
For $ x + 8=0 $, we get $ x=-8 $
# Answer:
The solutions are $ x = 0 $ and $ x=-8 $

### Problem 5: $ x^2-3x - 18=0 $
# Explanation:
## Step1: Factor the quadratic
We need two numbers that multiply to $ - 18$ and add up to $ - 3$. The numbers are $ - 6$ and $ 3$. So $ x^2-3x - 18=(x - 6)(x+3)=0 $
## Step2: Apply zero - product property
$ x - 6=0 $ or $ x + 3=0 $
## Step3: Solve for $ x $
For $ x - 6=0 $, $ x = 6 $; for $ x+3 = 0 $, $ x=-3 $
# Answer:
The solutions are $ x = 6 $ and $ x=-3 $

### Problem 6: $ 4x-x^2 + 8=0 $
# Explanation:
## Step1: Rewrite in standard form
$ -x^2+4x + 8=0 $, multiply both sides by $ - 1$ to get $ x^2-4x - 8=0 $
## Step2: Use quadratic formula $ x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}$, where $ a = 1$, $ b=-4 $, $ c=-8 $
First, calculate the discriminant $ \Delta=b^2-4ac=(-4)^2-4\times1\times(-8)=16 + 32=48 $
## Step3: Find the roots
$ x=\frac{4\pm\sqrt{48}}{2}=\frac{4\pm4\sqrt{3}}{2}=2\pm2\sqrt{3}$
$ \sqrt{3}\approx1.732 $, so $ 2 + 2\sqrt{3}\approx2 + 3.464 = 5.464$ and $ 2-2\sqrt{3}\approx2-3.464=-1.464 $
The root $ 2 - 2\sqrt{3}$ is between $ - 2$ and $ - 1$, and the root $ 2 + 2\sqrt{3}$ is between $ 5$ and $ 6$
# Answer:
The roots are between $ - 2$ and $ - 1$, $ 5$ and $ 6$ (exact roots: $ x = 2\pm2\sqrt{3}$)

### Problem 7: $ - 12-5x + 3x^2=0 $
# Explanation:
## Step1: Rewrite in standard form
$ 3x^2-5x - 12=0 $
## Step2: Factor the quadratic
We need two numbers that multiply to $ 3\times(-12)=-36$ and add up to $ - 5$. The numbers are $ - 9$ and $ 4$. So $ 3x^2-9x + 4x-12 = 0$, $ 3x(x - 3)+4(x - 3)=0$, $ (3x + 4)(x - 3)=0 $
## Step3: Apply zero - product property
$ 3x+4 = 0$ or $ x - 3=0 $
## Step4: Solve for $ x $
For $ 3x+4 = 0$, $ x=-\frac{4}{3}\approx - 1.33$; for $ x - 3=0$, $ x = 3$
The root $ -\frac{4}{3}$ is between $ - 2$ and $ - 1$
# Answer:
The solutions are $ x = 3$ and $ x=-\frac{4}{3}$ (the root $ -\frac{4}{3}$ is between $ - 2$ and $ - 1$)

### Problem 8: $ x^2-6x + 4=-8 $
# Explanation:
## Step1: Rewrite in standard form
$ x^2-6x+12 = 0 $
## Step2: Calculate the discriminant $ \Delta=b^2 - 4ac=(-6)^2-4\times1\times12=36 - 48=-12<0 $
Since the discriminant is negative, there are no real roots. But if we consider the function $ y=x^2-6x + 12=(x - 3)^2+3$, the minimum value of the function is $ 3>0$, so the graph of the function does not intersect the $ x$ - axis.
# Answer:
No real roots

### Problem 9: $ 9-x^2=12 $
# Explanation:
## Step1: Rewrite in standard form
$ -x^2-3 = 0$ or $ x^2=-3 $
Since the square of a real number is non - negative, there are no real roots.
# Answer:
No real roots

### Problem 10: $ 5x^2+10x - 4=-6 $
# Explanation:
## Step1: Rewrite in standard form
$ 5x^2+10x + 2=0 $
## Step2: Use quadratic formula $ x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}$, where $ a = 5$, $ b = 10$, $ c = 2$
The discriminant $ \Delta=b^2-4ac=10^2-4\times5\times2=100 - 40 = 60$
## Step3: Find the roots
$ x=\frac{-10\pm\sqrt{60}}{10}=\frac{-10\pm2\sqrt{15}}{10}=\frac{-5\pm\sqrt{15}}{5}=-1\pm\frac{\sqrt{15}}{5}$
$ \sqrt{15}\approx3.872 $, so $ -1+\frac{\sqrt{15}}{5}\approx-1 + 0.774=-0.226$ and $ -1-\frac{\sqrt{15}}{5}\approx-1 - 0.774=-1.774 $
The root $ -1-\frac{\sqrt{15}}{5}$ is between $ - 2$ and $ - 1$, and the root $ -1+\frac{\sqrt{15}}{5}$ is between $ - 1$ and $ 0$
# Answer:
The roots are between $ - 2$ and $ - 1$, $ - 1$ and $ 0$ (exact roots: $ x=-1\pm\frac{\sqrt{15}}{5}$)

### Problem 11: $ x^2-20=2 + x $
# Explanation:
## Step1: Rewrite in standard form
$ x^2-x-22 = 0 $
## Step2: Use quadratic formula $ x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}$, where $ a = 1$, $ b=-1 $, $ c=-22 $
The discriminant $ \Delta=b^2-4ac=(-1)^2-4\times1\times(-22)=1 + 88 = 89$
## Step3: Find the roots
$ x=\frac{1\pm\sqrt{89}}{2}$
$ \sqrt{89}\approx9.434 $, so $ \frac{1+\sqrt{89}}{2}\approx\frac{1 + 9.434}{2}\approx5.217$ and $ \frac{1-\sqrt{89}}{2}\approx\frac{1-9.434}{2}\approx - 4.217$
The root $ \frac{1-\sqrt{89}}{2}$ is between $ - 5$ and $ - 4$, and the root $ \frac{1+\sqrt{89}}{2}$ is between $ 5$ and $ 6$
# Answer:
The roots are between $ - 5$ and $ - 4$, $ 5$ and $ 6$ (exact roots: $ x=\frac{1\pm\sqrt{89}}{2}$)

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QUESTION IMAGE

Question

Explanation:

Problem 4: \( x^2 + 8x = 0 \)

Step1: Factor out \( x \)

Step2: Apply zero - product property

Step3: Solve for \( x \)

Step1: Factor the quadratic

Step2: Apply zero - product property

Step3: Solve for \( x \)

Step1: Rewrite in standard form

Step2: Use quadratic formula \( x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\), where \( a = 1\), \( b=-4 \), \( c=-8 \)

Step3: Find the roots

Answer:

Problem 5: \( x^2-3x - 18=0 \)