How do you solve using the completing the square method #5x^2 - 7x + 2 = 0#?
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To solve the quadratic equation 5x^2 - 7x + 2 = 0 using the completing the square method, follow these steps:
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Move the constant term to the other side of the equation: 5x^2 - 7x = -2
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Divide every term by the coefficient of x^2 (in this case, 5): x^2 - (7/5)x = -2/5
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Add and subtract the square of half of the coefficient of x: x^2 - (7/5)x + (49/100) = -2/5 + (49/100)
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Simplify the right side of the equation: x^2 - (7/5)x + (49/100) = -10/20 + 49/100 x^2 - (7/5)x + (49/100) = -50/100 + 49/100 x^2 - (7/5)x + (49/100) = -1/100
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Rewrite the left side as a squared binomial: (x - 7/10)^2 = -1/100
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Take the square root of both sides (note that there are two solutions): x - 7/10 = ±√(-1/100) x - 7/10 = ±(i/10)
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Solve for x by adding 7/10 to both sides: x = 7/10 ± (i/10)
The solutions to the quadratic equation 5x^2 - 7x + 2 = 0 using the completing the square method are: x = 7/10 + (i/10) and x = 7/10 - (i/10), where i is the imaginary unit (i^2 = -1).
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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