How can solve this using completing the squuare?

x^2 - 5a^2 - 4ax = 0

WHERE a IS CONSTANT

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- Sep 25th 2006, 05:27 AM^_^Engineer_Adam^_^Help !! completing the sqaure
How can solve this using completing the squuare?

x^2 - 5a^2 - 4ax = 0

WHERE a IS CONSTANT - Sep 25th 2006, 06:09 AMtopsquark
First:

x^2 - 4ax = 5a^2

Now, we want to add a constant to this such that the LHS is a perfect square:

(x + b)^2 = x^2 + 2bx + b^2

So comparing the linear term in this to the equation above I get that

-4a = 2b, thus b = -2a

So we wish to add b^2 = (-2a)^2 = 4a^2 to both sides:

x^2 - 4ax + 4a^2 = 5a^2 + 4a^2

(x - 2a)^2 = 9a^2

x - 2a = (+/-)3a

x = 2a (+/-) 3a

So x = 5a or x = -a.

-Dan - Sep 25th 2006, 06:12 AMSoroban
Hello, ^_^Engineer_Adam^_^!

Quote:

How can solve this using completing the square?

. . x² - 5a² - 4ax .= .0 . where*a*is constant.

The same way as we've been taught.

. . . . . . . . . . . . . . x² - 4ax .= .5a²

Then: . . . . . x² - 4ax + 4a² .= .5a² + 4a²

And: . . . . . . . . . . (x - 2a)² .= .9a²

Take square roots: . . x - 2a .= .±3a

So we have: . . . . . . . . . . x .= .2a ± 3a

Therefore: . x .= .5a, -a

Edit: Too fast for me, Dan! - Sep 26th 2006, 03:02 AM^_^Engineer_Adam^_^
thanks :D