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Solve

9x^2 + 42x + 49 = 0

User Yang Meyer
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1 Answer

4 votes

Because of the relatively large coefficients {9, 42, 49}, applying the quadratic formula would be a bit messy. Instead, I've chosen to "complete the square:"


9x^2 + 42x + 49 = 0 can be re-written as 9 [ x^2 + (42/9)x ] = -49


Dividing both sides by 9, we get [ x^2 + (42/9)x ] = - 49/9


Completing the square: [ x^2 + (42/9)x + (21/9)^2 - (21/9)^2 ] = -49/9

[ x + 21/9 ]^2 = 441/81 - 441/81 = 0


Then [ x + 21/9 ] = 0, and x = -21/9 (this is a double root).

User David Weng
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