help w/ quadratic?

Question

problem w/ quad. formula:

2 +-sqrt(4+4p^2) / 2

the discriminant is supposed to simplify into 1 + p^2, but i don't understand how. can some1 help clarify this pls?

Answer 1

It's all about simplifying the square root.

sqrt(4 + 4p^2)

Note that inside, we can factor out a 4.

sqrt(4(1 + p^2))

Now, we can decompose sqrt(ab) as sqrt(a)sqrt(b). Doing so for this, we get

sqrt(4)sqrt(1 + p^2)

And we know what the square root of 4 is, so we get

2 sqrt(1 + p^2)

Let's plug this back into what you currently have.

[2 +/- 2 sqrt(1 + p^2)] / 2

Notice all those 2s... we can divide each of the terms by 2 and cancel them out, giving us

1 +/- sqrt(1 + p^2)

Answer 2

Consider the discriminant which is:-

± [√(4p² + 4) ] = ± [√4(p² + 1)]= ± 2 √ (p² + 1)

Solution of quadratic then becomes :-

[2 ± 2 √(p² + 1)] / 2 = 1 ± √(p² + 1)

Answer 3

Factor out the 4:

4+4p² = 4(1+p²)

Apply √ab=√a√b:

√(4(1+p²)) = √4√(1+p²)
= 2√(1+p²)

So your expression becomes

2±2√(1+p²) / 2
=1±√(1+p²)

Answer 4

4 + 4p^2

factor out the 4

4(1 + p^2)

now you have
sqrt [4(1+p^2)]
= sqrt 4 * sqrt (1 + p^2)
= 2 * sqrt (1 + p^2)

All those twos cancel in the quad formula, so you're final answer is

+- sqrt (1 + p^2)