Question

question 9 of 10
which of the following are solutions to the equation below?
check all that apply.
$2x^2 + 18x = 20$

a. -2
b. -1
c. -2
d. 1
e. 20
f. -10

Explanation:

Step1: Rewrite the equation in standard form

First, we rewrite the given equation \(2x^{2}+18x = 20\) in standard quadratic form \(ax^{2}+bx + c=0\) by subtracting 20 from both sides:
\(2x^{2}+18x - 20=0\)
We can simplify this equation by dividing all terms by 2:
\(x^{2}+9x - 10=0\)

Step2: Factor the quadratic equation

We factor the quadratic equation \(x^{2}+9x - 10 = 0\). We need two numbers that multiply to \(- 10\) and add up to \(9\). The numbers are \(10\) and \(-1\) since \(10\times(-1)=-10\) and \(10+( - 1)=9\). So the factored form is:
\((x + 10)(x-1)=0\)

Step3: Solve for x

Using the zero - product property, if \(ab = 0\), then either \(a = 0\) or \(b = 0\).
For \((x + 10)(x - 1)=0\), we have two cases:
Case 1: \(x+10=0\), then \(x=-10\)
Case 2: \(x - 1=0\), then \(x = 1\)

We can also check the options by substituting the values into the original equation \(2x^{2}+18x=20\):

• For option A (\(x=-2\)):

Left - hand side (LHS) \(=2\times(-2)^{2}+18\times(-2)=2\times4-36 = 8 - 36=-28
eq20\)

• For option B (\(x = - 1\)):

LHS \(=2\times(-1)^{2}+18\times(-1)=2 - 18=-16
eq20\)

• For option D (\(x = 1\)):

LHS \(=2\times1^{2}+18\times1=2 + 18=20\), which is equal to the right - hand side (RHS)

• For option F (\(x=-10\)):

LHS \(=2\times(-10)^{2}+18\times(-10)=2\times100-180=200 - 180 = 20\), which is equal to the RHS

Answer:

D. 1, F. - 10