Question

consider the quadratic equation $x^2 - 10x = 11$
what are the solutions (if none, write none).
first x=?
second x=?
question 4
2 pts
solve for x:
$x^2 + 8x + 16 = 0$

Explanation:

First Problem: Solve \( x^2 - 10x = 11 \)

Step1: Rewrite in standard form

Subtract 11 from both sides: \( x^2 - 10x - 11 = 0 \)

Step2: Factor the quadratic

Find two numbers that multiply to -11 and add to -10. These numbers are -11 and 1. So, \( (x - 11)(x + 1) = 0 \)

Step3: Solve for \( x \)

Set each factor equal to zero:

• \( x - 11 = 0 \) gives \( x = 11 \)
• \( x + 1 = 0 \) gives \( x = -1 \)

Step1: Recognize the perfect square trinomial

The equation is \( (x + 4)^2 = 0 \) (since \( (a + b)^2 = a^2 + 2ab + b^2 \), here \( a = x \), \( b = 4 \), \( 2ab = 8x \), \( b^2 = 16 \))

Step2: Solve for \( x \)

Take the square root of both sides: \( x + 4 = 0 \), so \( x = -4 \) (double root)

Answer:

First \( x = 11 \)
Second \( x = -1 \)

Second Problem: Solve \( x^2 + 8x + 16 = 0 \)