Question

Solve for y: 4y2 4qy-(p2-q2)=0

Answer 1

Assuming you're talking about

$4 {y}^{2} + 4qy + ( {p}^{2} - {q}^{2} )$

Here, a = 4, b = 4q and c = (p^2-q^2)


Applying the quadratic formula for roots,

$\frac{ - b ± \sqrt{ {b}^{2} - 4ac } }{2a}$


Substitute the values,

$\frac{ - 4q± \sqrt{ {(4q)}^{2} - 4(4)( {p}^{2} - {q}^{2}) } }{2(4)} \\ \\ = \frac{ - 4q± \sqrt{16 {q}^{2} - 16 {p }^{2} + 16 {q}^{2} } }{8} \\ \\ = \frac{ - 4q± 4p\sqrt{ - 1 } }{8} \\ \\ = \frac{ - 4q±4pi}{8} \\ \\ = \frac{ - q±pi}{2}$

Therefore, the two values for y are these.

Answer 2

Answer:

There are two values for y:

$y=\frac{-q^2 + p i}{2}$; and

$y=\frac{-q^2 - p i}{2}$;

Explanation:

The given equation :

$4 y^2+4 q y+\left(p^2-q^2\right) = 0$

we can use the quadratic formula in order to find y:

So, In the given equation,

a = 4; b = 4q; and c= $\left(p^2-q^2\right)$

So, applying the quadratic formula, we get:

$y = \frac{-b^{2} \pm \sqrt{b^2-4 a c}}{2 a}$

$\begin{aligned}&y= \frac{-4 q ^{2} \pm \sqrt{(4 q)^2-4(4)\left(p^2-q^2\right)}}{2(4)} \\&y =\frac{-4 q ^{2} \pm \sqrt{16 q^2-16 p^2+16 q^2}}{8} \\& y=\frac{-4 q^{2} \pm 4 p \sqrt{-1}}{8} \\& y=\frac{-4 q^{2} \pm 4 p i}{8} \\&y =\frac{-q^{2} \pm p i}{2}\end{aligned}$

Therefore, y has two values:

$y=\frac{-q^2 + p i}{2}$;

$y=\frac{-q^2 - p i}{2}$

To learn more about quadratic formula, visit:

https://brainly.in/question/22194098

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https://brainly.in/question/9605334

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