To solve the quadratic equation \( x^2 - 18x = -80 \), let's go through each part step-by-step.
Part A: Finding c for a Perfect Square Trinomial
The equation can be rewritten as: \[ x^2 - 18x + c = -80 + c \]
To determine c such that \( x^2 - 18x + c \) is a perfect square trinomial, we need to complete the square.
The general form for completing the square for the equation \( x^2 - bx \) is: \[ c = \left(\frac{b}{2}\right)^2 \]
Here, \( b = 18 \): \[ c = \left(\frac{18}{2}\right)^2 = 9^2 = 81 \]
So, the value of c that makes the equation a perfect square trinomial is: Part A: c = 81
Part B: Factor the Perfect Square Trinomial
Now we can rewrite the equation including our value of c: \[ x^2 - 18x + 81 = -80 + 81 \] This simplifies to: \[ x^2 - 18x + 81 = 1 \]
The left-hand side factors as: \[ (x - 9)^2 = 1 \]
So we can rewrite the equation as: Part B: (x - 9)² = 1
Part C: Solve for x (lesser number)
To solve the equation \( (x - 9)^2 = 1 \), we take the square root of both sides: \[ x - 9 = \pm 1 \]
This leads to two solutions:
- \( x - 9 = 1 \) → \( x = 10 \)
- \( x - 9 = -1 \) → \( x = 8 \)
The lesser number from these solutions is: Part C: x = 8
Part D: Solve for x (greater number)
The greater number from the solutions derived above is: Part D: x = 10
Summary of Values
- Part A: c = 81
- Part B: (x - 9)² = 1
- Part C: x = 8
- Part D: x = 10