solving quadratics test review
solve by using square roots. simplify any radicals. no decimals.
1. $x^2 + 81 = 0$ 2. $2x^2 - 1 = 49$ 3. $5x^2 - 6 = -78$

solve.
4. $x(x + 1) = 0$ 5. $2(3x + 5)(x - 4) = 0$ 6. $3x(x - 2) = 0$

solve by factoring.
7. $4x^2 - 12x - 16 = 0$ 8. $2x^2 + 5x - 24 = -12$ 9. $4x^2 = 20x$

find the discriminant and describe the number and type of solutions. describe the number of x - intercepts.
10. $-x^2 + 4x - 13 = 0$ 11. $x^2 - 9 = 2x$ 12. $25x^2 - 8x = 12x - 4$

solve using the quadratic formula. simplify any radicals. no decimals.
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
13. $4x^2 - 12x - 16 = 0$ 14. $x^2 + 4x = 11$ 15. $4x^2 - 2x + 3 = x^2 + 2$

Question

solving quadratics test review
solve by using square roots. simplify any radicals. no decimals.
1. $x^2 + 81 = 0$  2. $2x^2 - 1 = 49$  3. $5x^2 - 6 = -78$

solve.
4. $x(x + 1) = 0$  5. $2(3x + 5)(x - 4) = 0$  6. $3x(x - 2) = 0$

solve by factoring.
7. $4x^2 - 12x - 16 = 0$  8. $2x^2 + 5x - 24 = -12$  9. $4x^2 = 20x$

find the discriminant and describe the number and type of solutions. describe the number of x - intercepts.
10. $-x^2 + 4x - 13 = 0$  11. $x^2 - 9 = 2x$  12. $25x^2 - 8x = 12x - 4$

solve using the quadratic formula. simplify any radicals. no decimals.
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
13. $4x^2 - 12x - 16 = 0$  14. $x^2 + 4x = 11$  15. $4x^2 - 2x + 3 = x^2 + 2$

Accepted Answer

### Problem 1: $ x^2 + 81 = 0 $
# Explanation:
## Step 1: Isolate $ x^2 $
Subtract 81 from both sides: $ x^2 = -81 $
## Step 2: Take square roots
Since the square of a real number can't be negative, in real numbers, no solution. In complex numbers, $ x = \pm \sqrt{-81} = \pm 9i $ (but the problem says "NO DECIMALS"—if we consider real numbers, no solution; if complex is allowed, $ \pm 9i $)
# Answer:
No real solution (or $ x = \pm 9i $ in complex numbers)

### Problem 2: $ 2x^2 - 1 = 49 $
# Explanation:
## Step 1: Isolate $ x^2 $
Add 1 to both sides: $ 2x^2 = 50 $
Divide by 2: $ x^2 = 25 $
## Step 2: Take square roots
$ x = \pm \sqrt{25} = \pm 5 $
# Answer:
$ x = 5 $ or $ x = -5 $

### Problem 3: $ 5x^2 - 6 = -78 $
# Explanation:
## Step 1: Isolate $ x^2 $
Add 6 to both sides: $ 5x^2 = -72 $
Divide by 5: $ x^2 = -\frac{72}{5} $
## Step 2: Take square roots
In real numbers, no solution. In complex numbers, $ x = \pm \sqrt{-\frac{72}{5}} = \pm \frac{6i\sqrt{10}}{5} $
# Answer:
No real solution (or $ x = \pm \frac{6i\sqrt{10}}{5} $ in complex numbers)

### Problem 4: $ x(x + 1) = 0 $
# Explanation:
## Step 1: Apply zero - product property
If $ ab = 0 $, then $ a = 0 $ or $ b = 0 $. So $ x = 0 $ or $ x + 1 = 0 $
## Step 2: Solve for x
For $ x + 1 = 0 $, subtract 1 from both sides: $ x=-1 $
# Answer:
$ x = 0 $ or $ x=-1 $

### Problem 5: $ 2(3x + 5)(x - 4)=0 $
# Explanation:
## Step 1: Apply zero - product property
Since $ 2
eq0 $, we have $ 3x + 5 = 0 $ or $ x - 4 = 0 $
## Step 2: Solve for x
For $ 3x+5 = 0 $, subtract 5 from both sides and divide by 3: $ 3x=-5\Rightarrow x =-\frac{5}{3} $
For $ x - 4 = 0 $, add 4 to both sides: $ x = 4 $
# Answer:
$ x=-\frac{5}{3} $ or $ x = 4 $

### Problem 6: $ 3x(x - 2)=0 $
# Explanation:
## Step 1: Apply zero - product property
We have $ 3x = 0 $ or $ x - 2 = 0 $
## Step 2: Solve for x
For $ 3x = 0 $, divide by 3: $ x = 0 $
For $ x - 2 = 0 $, add 2 to both sides: $ x = 2 $
# Answer:
$ x = 0 $ or $ x = 2 $

### Problem 7: $ 4x^2-12x - 16 = 0 $
# Explanation:
## Step 1: Factor out the GCF
The GCF of 4, - 12, - 16 is 4. So $ 4(x^2-3x - 4)=0 $
## Step 2: Factor the quadratic
Factor $ x^2-3x - 4 $. We need two numbers that multiply to - 4 and add to - 3. The numbers are - 4 and 1. So $ x^2-3x - 4=(x - 4)(x+1) $
The equation becomes $ 4(x - 4)(x + 1)=0 $
## Step 3: Apply zero - product property
Since $ 4
eq0 $, $ x - 4 = 0 $ or $ x + 1 = 0 $
## Step 4: Solve for x
For $ x - 4 = 0 $, $ x = 4 $; for $ x + 1 = 0 $, $ x=-1 $
# Answer:
$ x = 4 $ or $ x=-1 $

### Problem 8: $ 2x^2+5x - 24=-12 $
# Explanation:
## Step 1: Rewrite in standard form
Add 12 to both sides: $ 2x^2+5x - 12 = 0 $
## Step 2: Factor the quadratic
We need two numbers that multiply to $ 2\times(-12)=-24 $ and add to 5. The numbers are 8 and - 3.
Rewrite the middle term: $ 2x^2+8x-3x - 12 = 0 $
Factor by grouping: $ 2x(x + 4)-3(x + 4)=0\Rightarrow(2x - 3)(x + 4)=0 $
## Step 3: Apply zero - product property
$ 2x - 3 = 0 $ or $ x + 4 = 0 $
## Step 4: Solve for x
For $ 2x - 3 = 0 $, $ 2x=3\Rightarrow x=\frac{3}{2} $; for $ x + 4 = 0 $, $ x=-4 $
# Answer:
$ x=\frac{3}{2} $ or $ x=-4 $

### Problem 9: $ 4x^2 = 20x $
# Explanation:
## Step 1: Rewrite in standard form
Subtract $ 20x $ from both sides: $ 4x^2-20x = 0 $
## Step 2: Factor out the GCF
Factor out 4x: $ 4x(x - 5)=0 $
## Step 3: Apply zero - product property
$ 4x = 0 $ or $ x - 5 = 0 $
## Step 4: Solve for x
For $ 4x = 0 $, $ x = 0 $; for $ x - 5 = 0 $, $ x = 5 $
# Answer:
$ x = 0 $ or $ x = 5 $

### Problem 10: $ -x^2+4x - 13 = 0 $
# Explanation:
## Step 1: Identify a, b, c
For the quadratic equation $ ax^2+bx + c = 0 $, here $ a=-1 $, $ b = 4 $, $ c=-13 $
## Step 2: Calculate the discriminant $ D=b^2-4ac $
$ D=(4)^2-4\times(-1)\times(-13)=16 - 52=-36 $
## Step 3: Analyze the discriminant
Since $ D=-36<0 $, the equation has two complex conjugate solutions ($ x=\frac{-b\pm\sqrt{D}}{2a}=\frac{-4\pm\sqrt{-36}}{-2}=\frac{-4\pm6i}{-2}=2\mp3i $) and 0 x - intercepts (in real plane)
# Answer:
Discriminant: - 36; Two complex solutions; 0 x - intercepts

### Problem 11: $ x^2-9 = 2x $
# Explanation:
## Step 1: Rewrite in standard form
Subtract $ 2x $ and add 9 to both sides: $ x^2-2x - 9 = 0 $
## Step 2: Identify a, b, c
$ a = 1 $, $ b=-2 $, $ c=-9 $
## Step 3: Calculate the discriminant $ D=b^2-4ac $
$ D=(-2)^2-4\times1\times(-9)=4 + 36 = 40 $
## Step 4: Analyze the discriminant
Since $ D = 40>0 $ and not a perfect square, the equation has two distinct irrational real solutions ($ x=\frac{2\pm\sqrt{40}}{2}=\frac{2\pm2\sqrt{10}}{2}=1\pm\sqrt{10} $) and 2 x - intercepts
# Answer:
Discriminant: 40; Two distinct irrational real solutions; 2 x - intercepts

### Problem 12: $ 25x^2-8x = 12x - 4 $
# Explanation:
## Step 1: Rewrite in standard form
Subtract $ 12x $ and add 4 to both sides: $ 25x^2-20x + 4 = 0 $
## Step 2: Identify a, b, c
$ a = 25 $, $ b=-20 $, $ c = 4 $
## Step 3: Calculate the discriminant $ D=b^2-4ac $
$ D=(-20)^2-4\times25\times4=400 - 400 = 0 $
## Step 4: Analyze the discriminant
Since $ D = 0 $, the equation has one real solution (a repeated root) ($ x=\frac{-b\pm\sqrt{D}}{2a}=\frac{20\pm0}{50}=\frac{2}{5} $) and 1 x - intercept (a repeated root)
# Answer:
Discriminant: 0; One real solution (repeated); 1 x - intercept

### Problem 13: $ 4x^2-12x - 16 = 0 $
# Explanation:
## Step 1: Identify a, b, c
For $ 4x^2-12x - 16 = 0 $, $ a = 4 $, $ b=-12 $, $ c=-16 $
## Step 2: Apply quadratic formula $ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} $
First, calculate discriminant $ D=(-12)^2-4\times4\times(-16)=144 + 256 = 400 $
$ x=\frac{12\pm\sqrt{400}}{8}=\frac{12\pm20}{8} $
## Step 3: Find the two solutions
For the plus sign: $ x=\frac{12 + 20}{8}=\frac{32}{8}=4 $
For the minus sign: $ x=\frac{12-20}{8}=\frac{-8}{8}=-1 $
# Answer:
$ x = 4 $ or $ x=-1 $

### Problem 14: $ x^2+4x = 11 $
# Explanation:
## Step 1: Rewrite in standard form
Subtract 11 from both sides: $ x^2+4x - 11 = 0 $
## Step 2: Identify a, b, c
$ a = 1 $, $ b = 4 $, $ c=-11 $
## Step 3: Apply quadratic formula
Discriminant $ D=(4)^2-4\times1\times(-11)=16 + 44 = 60 $
$ x=\frac{-4\pm\sqrt{60}}{2}=\frac{-4\pm2\sqrt{15}}{2}=-2\pm\sqrt{15} $
# Answer:
$ x=-2+\sqrt{15} $ or $ x=-2-\sqrt{15} $

### Problem 15: $ 4x^2-2x + 3 = x^2+2 $
# Explanation:
## Step 1: Rewrite in standard form
Subtract $ x^2 $ and 2 from both sides: $ 3x^2-2x + 1 = 0 $
## Step 2: Identify a, b, c
$ a = 3 $, $ b=-2 $, $ c = 1 $
## Step 3: Apply quadratic formula
Discriminant $ D=(-2)^2-4\times3\times1=4 - 12=-8 $
$ x=\frac{2\pm\sqrt{-8}}{6}=\frac{2\pm2\sqrt{2}i}{6}=\frac{1\pm\sqrt{2}i}{3} $
# Answer:
$ x=\frac{1+\sqrt{2}i}{3} $ or $ x=\frac{1-\sqrt{2}i}{3} $ (two complex solutions)

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QUESTION IMAGE

Question

Explanation:

Problem 1: \( x^2 + 81 = 0 \)

Step 1: Isolate \( x^2 \)

Step 2: Take square roots

Step 1: Isolate \( x^2 \)

Step 2: Take square roots

Step 1: Isolate \( x^2 \)

Step 2: Take square roots

Answer:

Problem 2: \( 2x^2 - 1 = 49 \)