Question

(x+8)^2 = 9
(x+6)^2= -7
topic 5: solve by using the quadratic formula

1. p^2 + 14p - 38 = 0
2. v^2 + 6v - 59 = 0
3. a^2 + 14a - 51 = 0
4. x^2 - 12x + 11 = 0
5. x^2 + 6x + 8 = 0
6. n^2 - 2n - 3 = 0

quadratic formula:
for ax^2 + bx + c = 0,
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
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Explanation:

Let's solve problem 1: \( p^2 + 14p - 38 = 0 \) using the quadratic formula.

Step 1: Identify \( a \), \( b \), \( c \)

For the equation \( p^2 + 14p - 38 = 0 \), we have \( a = 1 \), \( b = 14 \), \( c = -38 \).

Step 2: Substitute into the quadratic formula

The quadratic formula is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substituting the values, we get:
\( p = \frac{-14 \pm \sqrt{14^2 - 4(1)(-38)}}{2(1)} \)

Step 3: Calculate the discriminant

First, calculate \( b^2 - 4ac \):
\( 14^2 - 4(1)(-38) = 196 + 152 = 348 \)

Step 4: Simplify the square root

\( \sqrt{348} = \sqrt{4 \times 87} = 2\sqrt{87} \)

Step 5: Solve for \( p \)

\( p = \frac{-14 \pm 2\sqrt{87}}{2} = -7 \pm \sqrt{87} \)

Now, let's solve problem 2: \( v^2 + 6v - 59 = 0 \)

Step 1: Identify \( a \), \( b \), \( c \)

Here, \( a = 1 \), \( b = 6 \), \( c = -59 \)

Step 2: Substitute into the quadratic formula

\( v = \frac{-6 \pm \sqrt{6^2 - 4(1)(-59)}}{2(1)} \)

Step 3: Calculate the discriminant

\( 6^2 - 4(1)(-59) = 36 + 236 = 272 \)

Step 4: Simplify the square root

\( \sqrt{272} = \sqrt{16 \times 17} = 4\sqrt{17} \)

Step 5: Solve for \( v \)

\( v = \frac{-6 \pm 4\sqrt{17}}{2} = -3 \pm 2\sqrt{17} \)

For problem 3: \( a^2 + 14a - 51 = 0 \)

Step 1: Identify \( a \), \( b \), \( c \)

\( a = 1 \), \( b = 14 \), \( c = -51 \)

Step 2: Substitute into the quadratic formula

\( a = \frac{-14 \pm \sqrt{14^2 - 4(1)(-51)}}{2(1)} \)

Step 3: Calculate the discriminant

\( 14^2 - 4(1)(-51) = 196 + 204 = 400 \)

Step 4: Simplify the square root

\( \sqrt{400} = 20 \)

Step 5: Solve for \( a \)

\( a = \frac{-14 \pm 20}{2} \)
So, \( a = \frac{-14 + 20}{2} = 3 \) or \( a = \frac{-14 - 20}{2} = -17 \)

Problem 4: \( x^2 - 12x + 11 = 0 \)

Step 1: Identify \( a \), \( b \), \( c \)

\( a = 1 \), \( b = -12 \), \( c = 11 \)

Step 2: Substitute into the quadratic formula

\( x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(1)(11)}}{2(1)} \)

Step 3: Calculate the discriminant

\( (-12)^2 - 4(1)(11) = 144 - 44 = 100 \)

Step 4: Simplify the square root

\( \sqrt{100} = 10 \)

Step 5: Solve for \( x \)

\( x = \frac{12 \pm 10}{2} \)
So, \( x = \frac{12 + 10}{2} = 11 \) or \( x = \frac{12 - 10}{2} = 1 \)

Problem 5: \( x^2 + 6x + 8 = 0 \)

Step 1: Identify \( a \), \( b \), \( c \)

\( a = 1 \), \( b = 6 \), \( c = 8 \)

Step 2: Substitute into the quadratic formula

\( x = \frac{-6 \pm \sqrt{6^2 - 4(1)(8)}}{2(1)} \)

Step 3: Calculate the discriminant

\( 6^2 - 4(1)(8) = 36 - 32 = 4 \)

Step 4: Simplify the square root

\( \sqrt{4} = 2 \)

Step 5: Solve for \( x \)

\( x = \frac{-6 \pm 2}{2} \)
So, \( x = \frac{-6 + 2}{2} = -2 \) or \( x = \frac{-6 - 2}{2} = -4 \)

Problem 6: \( n^2 - 2n - 3 = 0 \)

Step 1: Identify \( a \), \( b \), \( c \)

\( a = 1 \), \( b = -2 \), \( c = -3 \)

Step 2: Substitute into the quadratic formula

\( n = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-3)}}{2(1)} \)

Step 3: Calculate the discriminant

\( (-2)^2 - 4(1)(-3) = 4 + 12 = 16 \)

Step 4: Simplify the square root

\( \sqrt{16} = 4 \)

Step 5: Solve for \( n \)

\( n = \frac{2 \pm 4}{2} \)
So, \( n = \frac{2 + 4}{2} = 3 \) or \( n = \frac{2 - 4}{2} = -1 \)

Answer:

s:

1. \( p = -7 \pm \sqrt{87} \)
2. \( v = -3 \pm 2\sqrt{17} \)
3. \( a = 3 \) or \( a = -17 \)
4. \( x = 11 \) or \( x = 1 \)
5. \( x = -2 \) or \( x = -4 \)
6. \( n = 3 \) or \( n = -1 \)