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Ch. 1 - Equations and Inequalities
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514당신이 사용하는 게 아니라요?교과서 변경
2장, 문제 55

Solve each equation in Exercises 47–64 by completing the square. x25x+6=0x^2 - 5x + 6 = 0

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1
Start with the given quadratic equation: \(x^2 - 5x + 6 = 0\).
Move the constant term to the other side to isolate the quadratic and linear terms: \(x^2 - 5x = -6\).
To complete the square, take half of the coefficient of \(x\), which is \(-5\), divide by 2 to get \(-\frac{5}{2}\), then square it to get \(\left(-\frac{5}{2}\right)^2 = \frac{25}{4}\).
Add \(\frac{25}{4}\) to both sides of the equation to maintain equality: \(x^2 - 5x + \frac{25}{4} = -6 + \frac{25}{4}\).
Rewrite the left side as a perfect square trinomial: \(\left(x - \frac{5}{2}\right)^2 = -6 + \frac{25}{4}\). From here, you can simplify the right side and solve for \(x\) by taking the square root of both sides.

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주요 개념

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Solving Equations Using Square Roots

Once a quadratic equation is written as a perfect square equal to a constant, you can solve it by taking the square root of both sides. This step introduces both positive and negative roots, which are critical for finding all solutions.
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