How can solve this using completing the squuare?
x^2 - 5a^2 - 4ax = 0
WHERE a IS CONSTANT
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
Hello, ^_^Engineer_Adam^_^!
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!