Question

1. solve the equation by graphing.

$(x - 4)^2 + 3 = (-x + 5)(x - 1) + 12$

a. $square$ $x = -3$

b. $square$ $x \approx -1.98$

c. $square$ $x = 1$

d. $square$ $x = 4$

e. $square$ $x = 6$

f. $square$ $x = 2.5$

g. $square$ $x \approx 5.64$

h. $square$ no real solution

Explanation:

Step1: Simplify both sides

First, expand the right - hand side: \((-x + 5)(x - 1)+12=-x(x - 1)+5(x - 1)+12=-x^{2}+x + 5x-5 + 12=-x^{2}+6x + 7\)
The left - hand side is \((x - 4)^{2}+3=x^{2}-8x + 16+3=x^{2}-8x + 19\)
So the equation becomes \(x^{2}-8x + 19=-x^{2}+6x + 7\)

Step2: Rearrange the equation

Move all terms to the left - hand side: \(x^{2}-8x + 19+x^{2}-6x - 7 = 0\)
Combine like terms: \(2x^{2}-14x + 12 = 0\)
Divide through by 2: \(x^{2}-7x + 6 = 0\)

Step3: Solve the quadratic equation

Factor the quadratic: \(x^{2}-7x + 6=(x - 1)(x - 6)=0\)
Set each factor equal to zero: \(x-1 = 0\) or \(x - 6=0\)
So \(x = 1\) or \(x = 6\)

We can also check by substituting the values into the original equation:

• For \(x = 1\):

Left - hand side: \((1 - 4)^{2}+3=(-3)^{2}+3=9 + 3=12\)
Right - hand side: \((-1 + 5)(1 - 1)+12=(4)\times(0)+12 = 12\)

• For \(x = 6\):

Left - hand side: \((6 - 4)^{2}+3=(2)^{2}+3=4 + 3=7\)
Right - hand side: \((-6 + 5)(6 - 1)+12=(-1)\times(5)+12=-5 + 12=7\)

Answer:

c. \(x = 1\), e. \(x = 6\)