Roots with a difference!

After posting this problem I spent some of my time on the difference of zeroes of quadratic polynomial and came to the following result:

Consider a polynomial $f(x)=x^{2}-bx+a$ and suppose it's roots are $p$ and $q$ and you know the value $k$ where $p-q=k$ . Then $b^{2}-4a=k^{2}$

Proof:

$f(x)=x^{2}-bx+a$

From Vieta's formula

$p+q=b$

$and$

$pq=a$

as, $p-q=k$ therefore $p=q+k$

putting this in above equations

$2q+k=b$

$and$

$q^{2}+qk=a$

$b^{2}-4a=(2q+k)^{2}-a(q^{2}+qk)=4q^{2}+4qk+k^{2}-4q^{2}-4qk=\boxed{k^{2}}$

#Algebra

Easy Math Editor

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`2 \times 3`	$2 \times 3$
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Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$