To solve the problem using the quadratic formula, we'll work through each part step by step.
Part A: GCF
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Finding the GCF: The given quadratic is \( 2x^2 + 22x + 60 \). The GCF of the coefficients (2, 22, and 60) is 2.
The GCF is = 2
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Factoring out the GCF: We can factor out the GCF from the quadratic expression: \[ 2(x^2 + 11x + 30) \] The quadratic with the factored out the GCF is: \( 2(x^2 + 11x + 30) \)
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Values of a, b, and c: After factoring out the GCF, the coefficients for the quadratic \( x^2 + 11x + 30 \) are:
- \( a = 1 \)
- \( b = 11 \)
- \( c = 30 \)
Part B: Use the quadratic formula
We will now calculate using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Plugging in the values from Part A:
- \( a = 1 \)
- \( b = 11 \)
- \( c = 30 \)
The formula becomes: \[ x = \frac{-11 \pm \sqrt{11^2 - 4(1)(30)}}{2(1)} \]
We compute: \[ 11^2 = 121 \] \[ 4(1)(30) = 120 \] So, \[ x = \frac{-11 \pm \sqrt{121 - 120}}{2} \] Thus, \[ x = \frac{-11 \pm \sqrt{1}}{2} \]
Part C: Solve for x
Now, we can simplify the expression: \[ \sqrt{1} = 1 \]
So our expression for \( x \) is: \[ x = \frac{-11 \pm 1}{2} \]
Calculating the two solutions:
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For the lesser number: \[ x = \frac{-11 - 1}{2} = \frac{-12}{2} = -6 \]
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For the greater number: \[ x = \frac{-11 + 1}{2} = \frac{-10}{2} = -5 \]
Summary of Answers
Part A:
- GCF is = 2
- The quadratic with the factored out the GCF is: \( 2(x^2 + 11x + 30) \)
- \( a = 1 \)
- \( b = 11 \)
- \( c = 30 \)
Part B:
- Use the quadratic formula: \[ -11 \pm 1^2 - 4(1)(30) \] \[ \frac{}{} (2(1)) \]
Part C:
- Solve for x:
- \( x = -6 \) (lesser number)
- \( x = -5 \) (greater number)