Question

question 10 of 10
which of the following are solutions to the equation below?
check all that apply.
$x^2 + 4x - 9 = x + 1$

a. -4
b. -3
c. 2
d. 4
e. -5
f. -7

Explanation:

Step1: Rearrange the equation

First, we need to rearrange the given equation \(x^{2}+4x - 9=x + 1\) to the standard quadratic form \(ax^{2}+bx + c = 0\). Subtract \(x\) and \(1\) from both sides:
\(x^{2}+4x - x-9 - 1=0\)
Simplify the like terms:
\(x^{2}+3x - 10 = 0\)

Step2: Factor the quadratic equation

We factor the quadratic equation \(x^{2}+3x - 10 = 0\). We need two numbers that multiply to \(- 10\) and add up to \(3\). The numbers are \(5\) and \(-2\) since \(5\times(-2)=-10\) and \(5+(-2) = 3\). So we can factor the equation as:
\((x + 5)(x-2)=0\)

Step3: Solve for \(x\)

Using the zero - product property, if \(ab = 0\), then either \(a = 0\) or \(b = 0\). So we set each factor equal to zero:

• For \(x + 5=0\), we get \(x=-5\).
• For \(x - 2=0\), we get \(x = 2\).

We can also check the options by substituting the values of \(x\) into the original equation:

• For option A (\(x=-4\)):

Left - hand side (LHS): \((-4)^{2}+4\times(-4)-9=16-16 - 9=-9\)
Right - hand side (RHS): \(-4 + 1=-3\)
Since \(-9
eq-3\), \(x = - 4\) is not a solution.

• For option B (\(x=-3\)):

LHS: \((-3)^{2}+4\times(-3)-9=9-12 - 9=-12\)
RHS: \(-3 + 1=-2\)
Since \(-12
eq-2\), \(x=-3\) is not a solution.

• For option C (\(x = 2\)):

LHS: \(2^{2}+4\times2-9=4 + 8-9 = 3\)
RHS: \(2 + 1=3\)
Since \(LHS = RHS\), \(x = 2\) is a solution.

• For option D (\(x = 4\)):

LHS: \(4^{2}+4\times4-9=16 + 16-9=23\)
RHS: \(4 + 1=5\)
Since \(23
eq5\), \(x = 4\) is not a solution.

• For option E (\(x=-5\)):

LHS: \((-5)^{2}+4\times(-5)-9=25-20 - 9=-4\)
RHS: \(-5 + 1=-4\)
Since \(LHS = RHS\), \(x=-5\) is a solution.

• For option F (\(x=-7\)):

LHS: \((-7)^{2}+4\times(-7)-9=49-28 - 9=12\)
RHS: \(-7 + 1=-6\)
Since \(12
eq-6\), \(x=-7\) is not a solution.

Answer:

C. 2, E. -5