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If [tex][tex]$2x^2 - 21x + 27 = (2x - 3)(x - 9)$[/tex][/tex], which equation(s) should be solved to find the roots of [tex][tex]$2x^2 - 21x + 27 = 0$[/tex][/tex]? Check all that apply.

A. [tex][tex]$x - 9 = 0$[/tex][/tex]
B. [tex][tex]$x \div 9 = 0$[/tex][/tex]
C. [tex][tex]$2x - 3 = x - 9$[/tex][/tex]
D. [tex][tex]$2x + 3 = 0$[/tex][/tex]
E. [tex][tex]$2x - 3 = 0$[/tex][/tex]



Answer :

GinnyAnswer
To find the roots of the quadratic equation [tex]\(2x^2 - 21x + 27 = 0\)[/tex], we need to solve the factored form of the equation. Given that:
[tex]\[ 2x^2 - 21x + 27 = (2x - 3)(x - 9) = 0 \][/tex]

We need to find the values of [tex]\(x\)[/tex] that make the factored equation equal to zero. This can be done by setting each factor equal to zero and solving for [tex]\(x\)[/tex].

Let's consider each term in the factored form separately:

1. [tex]\(2x - 3 = 0\)[/tex]
2. [tex]\(x - 9 = 0\)[/tex]

### Solving these equations:

1. Solving [tex]\(2x - 3 = 0\)[/tex]:
[tex]\[ 2x - 3 = 0 \][/tex]
Add 3 to both sides:
[tex]\[ 2x = 3 \][/tex]
Now, divide by 2:
[tex]\[ x = \frac{3}{2} \][/tex]

2. Solving [tex]\(x - 9 = 0\)[/tex]:
[tex]\[ x - 9 = 0 \][/tex]
Add 9 to both sides:
[tex]\[ x = 9 \][/tex]

So, the roots of the original quadratic equation are [tex]\(x = \frac{3}{2}\)[/tex] and [tex]\(x = 9\)[/tex].

### Conclusion:
The equations that need to be solved to find these roots are:
- [tex]\(x - 9 = 0\)[/tex]
- [tex]\(2x - 3 = 0\)[/tex]

Thus, the correct answer is:
A. [tex]\(x - 9 = 0\)[/tex]
E. [tex]\(2x - 3 = 0\)[/tex]

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