To solve the quadratic equation \( 4x^2 + 12x + 5 = 0 \) by factoring by grouping, we will follow these steps:
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Identify the equation: \[ 4x^2 + 12x + 5 = 0 \]
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Multiply \(a\) and \(c\) (the coefficients of \(x^2\) and the constant term):
- Here, \( a = 4 \) and \( c = 5 \).
- Therefore, \( ac = 4 \times 5 = 20 \).
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Find two numbers that multiply to \(ac\) and add up to \(b\) (the coefficient of \(x\)):
- We need two numbers that multiply to 20 and add up to 12.
- These numbers are 10 and 2 because \( 10 \times 2 = 20 \) and \( 10 + 2 = 12 \).
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Rewrite the middle term using these numbers: \[ 4x^2 + 10x + 2x + 5 = 0 \]
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Group the terms: \[ (4x^2 + 10x) + (2x + 5) = 0 \]
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Factor out common factors from each group:
- From the first group \(4x^2 + 10x\), we can factor out \(2x\): \[ 2x(2x + 5) \]
- From the second group \(2x + 5\), there is no common factor: \[ 1(2x + 5) \]
- So now we have: \[ 2x(2x + 5) + 1(2x + 5) = 0 \]
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Factor out the common binomial factor: \[ (2x + 5)(2x + 1) = 0 \]
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Set each factor to zero: \[ 2x + 5 = 0 \quad \text{or} \quad 2x + 1 = 0 \]
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Solve for \(x\):
- From \(2x + 5 = 0\): \[ 2x = -5 \implies x = -\frac{5}{2} \]
- From \(2x + 1 = 0\): \[ 2x = -1 \implies x = -\frac{1}{2} \]
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Final solutions: \[ x = -\frac{5}{2} \quad \text{and} \quad x = -\frac{1}{2} \]
Thus, the solutions to the equation \( 4x^2 + 12x + 5 = 0 \) are \( x = -\frac{5}{2} \) and \( x = -\frac{1}{2} \).