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Question
Solve by completing the square:
x2 − 2ax + 3x − 6a = 0
Sum
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Solution
Given:
x2 − 2ax + 3x − 6a = 0
Complete the square:
`x^2 + (3 − 2a)x = 6a `
Add `[(3 − 2a)/2]^2` to both sides:
`(x + (3 − 2a)/2)^2 = 6a + (3 − 2a)^2/4`
Compute the right-hand side:
6`a + (9 − 12a + 4a^2)/4 = (4a^2 + 12a + 9)/4 = ((2a + 3)/2)^2`
So `(x + (3 − 2a)/2)^2 = ((2a + 3)/2)^2`
Take square roots:
`x + (3 − 2a)/2 = ±(2a + 3)/2`
x = 2a or x = −3
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