Question

solve by using the graphing method

1. $y = -x^2 + 2x + 1$ when $y = 0$

points ____ and ____
$x = $ ____ and $x = $ ____

1. $y = -2x^2 - 16x - 30$, for $y = -6$

points ____ and ____
$x = $ ____ and $x = $ ____

1. $4 = x^2 + 8x + 16$

points ____ and ____
$x = $ ____ and $x = $ ____

solve using the square root method.

1. $y = 4x^2 - 80$

1. $5(x + 2)^2 - 6 = 174$

1. $32 = 2(x - 8)^2$

1. $2x^2 + 30 = 6x^2 - 98$

1. $3x^2 - 72 = y$

1. $8(x - 1)^2 + 12 = 108$

Explanation:

Step1: Solve \( y = 4x^2 - 80 \) for \( x \) (assuming \( y = 0 \) as it's a quadratic solution context)

Set \( y = 0 \), so \( 0 = 4x^2 - 80 \). Add 80 to both sides: \( 4x^2 = 80 \). Divide by 4: \( x^2 = 20 \). Take square roots: \( x = \pm\sqrt{20} = \pm2\sqrt{5} \approx \pm4.47 \). But if we consider solving for \( x \) in terms of \( y \), rearrange: \( 4x^2 = y + 80 \), \( x^2 = \frac{y + 80}{4} \), \( x = \pm\sqrt{\frac{y + 80}{4}} = \pm\frac{\sqrt{y + 80}}{2} \). But likely \( y = 0 \) is intended. So \( x = \pm2\sqrt{5} \) or \( \pm4.47 \). Wait, maybe the problem is to solve for \( x \) when \( y = 0 \). So step by step:

1. Start with \( y = 4x^2 - 80 \).
2. Set \( y = 0 \): \( 0 = 4x^2 - 80 \).
3. Add 80 to both sides: \( 4x^2 = 80 \).
4. Divide both sides by 4: \( x^2 = 20 \).
5. Take square root of both sides: \( x = \pm\sqrt{20} = \pm2\sqrt{5} \approx \pm4.47 \).

Step2: Solve \( 5(x + 2)^2 - 6 = 174 \)

1. Add 6 to both sides: \( 5(x + 2)^2 = 180 \).
2. Divide by 5: \( (x + 2)^2 = 36 \).
3. Take square roots: \( x + 2 = \pm6 \).
4. Solve for \( x \): \( x = -2 \pm 6 \). So \( x = 4 \) or \( x = -8 \).

Step3: Solve \( 32 = 2(x - 8)^2 \)

1. Divide both sides by 2: \( 16 = (x - 8)^2 \).
2. Take square roots: \( x - 8 = \pm4 \).
3. Solve for \( x \): \( x = 8 \pm 4 \). So \( x = 12 \) or \( x = 4 \).

Step4: Solve \( 2x^2 + 30 = 6x^2 - 98 \)

1. Subtract \( 2x^2 \) from both sides: \( 30 = 4x^2 - 98 \).
2. Add 98 to both sides: \( 128 = 4x^2 \).
3. Divide by 4: \( 32 = x^2 \).
4. Take square roots: \( x = \pm\sqrt{32} = \pm4\sqrt{2} \approx \pm5.66 \).

Step5: Solve \( 3x^2 - 72 = y \) for \( x \) (assuming \( y = 0 \))

1. Set \( y = 0 \): \( 3x^2 - 72 = 0 \).
2. Add 72: \( 3x^2 = 72 \).
3. Divide by 3: \( x^2 = 24 \).
4. Take square roots: \( x = \pm\sqrt{24} = \pm2\sqrt{6} \approx \pm4.90 \).

Step6: Solve \( 8(x - 1)^2 + 12 = 108 \)

1. Subtract 12: \( 8(x - 1)^2 = 96 \).
2. Divide by 8: \( (x - 1)^2 = 12 \).
3. Take square roots: \( x - 1 = \pm\sqrt{12} = \pm2\sqrt{3} \approx \pm3.46 \).
4. Solve for \( x \): \( x = 1 \pm 2\sqrt{3} \approx 1 \pm 3.46 \), so \( x \approx 4.46 \) or \( x \approx -2.46 \).

Answer:

For problem 4 (\( y = 4x^2 - 80 \), \( y = 0 \)): \( x = \pm2\sqrt{5} \) (or \( \pm4.47 \))
For problem 5: \( x = 4 \) or \( x = -8 \)
For problem 6: \( x = 12 \) or \( x = 4 \)
For problem 7: \( x = \pm4\sqrt{2} \) (or \( \pm5.66 \))
For problem 8 (\( 3x^2 - 72 = y \), \( y = 0 \)): \( x = \pm2\sqrt{6} \) (or \( \pm4.90 \))
For problem 9: \( x = 1 \pm 2\sqrt{3} \) (or \( \approx4.46 \), \( \approx -2.46 \))

(Note: The original problems might have specific contexts, but the above are solutions using square root method as required.)