how to solve this quadratic equation by completing perfect square
a^2.u^2+2abcd.u-(1+2c)b^2.d^2=0

Question

how to solve this quadratic equation by completing perfect square
a^2.u^2+2abcd.u-(1+2c)b^2.d^2=0

anonymous · Answer

$$a^2u^2+2abcd.u-(1+2c)b^2d^2=0$$

mathstudent55 · Answer

The variable is u, and all other letters are constants?

anonymous · Answer

i did it till
$$(u+\frac{ bcd }{ a })^2=\frac{ (1+2c)b^2d^2 }{ a^2 }+(\frac{ bcd }{ a })^2$$

mathstudent55 · Answer

Let me try.
To compete the square, the leading coefficient must be 1, so if it isn't 1, we divide the entire equation by the leading coefficient.

mathstudent55 · Answer

$\large a^2u^2+2abcdu-(1+2c)b^2d^2=0$

$\large u^2+\dfrac{2abcdu}{a^2}-\dfrac{(1+2c)b^2d^2}{a^2}=0 $

$\large u^2+\dfrac{2bcdu}{a}-\dfrac{(1+2c)b^2d^2}{a^2}=0 $

mathstudent55 · Answer

Now we move the constant term to the right side:

$\large u^2+\dfrac{2bcdu}{a} = \dfrac{(1+2c)b^2d^2}{a^2}$

mathstudent55 · Answer

Now we add the square of half of the second term's coefficient to both sides.

$\large u^2+\dfrac{2bcd}{a}u + \left(\dfrac{bcd}{a}\right)^2 = \dfrac{(1+2c)b^2d^2}{a^2} + \left(\dfrac{bcd}{a}\right)^2 $

anonymous · Answer

i have written the next step

mathstudent55 · Answer

$\large \left(u +\dfrac{bcd}{a}\right)^2 = \dfrac{(1+2c)b^2d^2 +b^2c^2d^2}{a^2} $

mathstudent55 · Answer

So far we have the same.

mathstudent55 · Answer

Now take roots and don't forget +/- on the right side.
I am assuming the constants are nonnegative.

$\large u +\dfrac{bcd}{a} = \pm \sqrt{\dfrac{(1+2c)b^2d^2 +b^2c^2d^2}{a^2}}$

$\large u +\dfrac{bcd}{a} = \pm \dfrac{bd}{a}\sqrt{(1+2c) +c^2}$

anonymous · Answer

$$\sqrt{(1+2c)+c^2}=\sqrt{(c+1)^2}$$

anonymous · Answer

so
$$u+\frac{ bcd }{ a }=\pm[\frac{ bcd }{ a }+\frac{ bd }{ a }]$$

anonymous · Answer

so i think the answer is 
$$u=\frac{ bd }{ a }$$
and
$$u=\frac{ -bd(2c+1) }{ a }$$

mathstudent55 · Answer

$\large u =-\dfrac{bcd}{a} \pm \dfrac{bd}{a}\sqrt{(1+2c) +c^2}$

$\large u =\dfrac{-bcd \pm bd\sqrt{c^2 + 2c + 1}}{a}$

$\large u =\dfrac{-bdc \pm bd\sqrt{(c + 1)^2}}{a}$

$\large u =\dfrac{-bdc \pm bd(c + 1)}{a}$

$\large u =\dfrac{-bdc \pm (bdc + bd)}{a}$

$\large u =\dfrac{bd}{a}$   or   $\large u =\dfrac{-2bdc - bd}{a}$

$\large u =\dfrac{bd}{a}$   or   $\large u =\dfrac{-bd(2c + 1)}{a}$

mathstudent55 · Answer

Yes, you are correct.

anonymous · Answer

THANK YOU

mathstudent55 · Answer

You're welcome.