Question

determine the exact solutions
a. ( x^2 + 10x - 24 = 0 )
b. ( 5x^2 + 13 - 6 = 0 ) (note: likely a typo, maybe ( 5x^2 + 13x - 6 = 0 ) or similar)
c. ( (x - 3)^2 = 16 )

Explanation:

Part A: Solve \( x^2 + 10x - 24 = 0 \)

Step 1: Factor the quadratic

We need two numbers that multiply to \( -24 \) and add to \( 10 \). The numbers are \( 12 \) and \( -2 \). So, \( x^2 + 10x - 24=(x + 12)(x - 2)=0 \)

Step 2: Solve for \( x \)

Set each factor equal to zero:

• \( x + 12 = 0 \) gives \( x=-12 \)
• \( x - 2 = 0 \) gives \( x = 2 \)

Part B: Solve \( 5x^2+13 - 6 = 0 \) (assuming it's \( 5x^2+13x - 6 = 0 \) due to possible typo)

Step 1: Factor the quadratic

We need two numbers that multiply to \( 5\times(-6)=-30 \) and add to \( 13 \). The numbers are \( 15 \) and \( -2 \). Rewrite the middle term: \( 5x^2+15x - 2x - 6 = 0 \)
Factor by grouping: \( 5x(x + 3)-2(x + 3)=0\Rightarrow(5x - 2)(x + 3)=0 \)

Step 2: Solve for \( x \)

• \( 5x - 2 = 0 \) gives \( x=\frac{2}{5} \)
• \( x + 3 = 0 \) gives \( x=-3 \)

Part C: Solve \( (x - 3)^2=16 \)

Step 1: Take square roots

Take the square root of both sides: \( x - 3=\pm\sqrt{16}=\pm4 \)

Step 2: Solve for \( x \)

• When \( x - 3 = 4 \), \( x=4 + 3=7 \)
• When \( x - 3=-4 \), \( x=-4 + 3=-1 \)

Answer:

• Part A: \( x=-12, x = 2 \)
• Part B: \( x=\frac{2}{5}, x=-3 \) (assuming corrected equation)
• Part C: \( x = 7, x=-1 \)