# 1.Show roots of x^2+2x-(m^2-1)=0 are rational values of m. 2.The equation in terms of x, 2q(p+r)x+(p^2+q^2)x^2+(q^2+r^2)=0 has real roots.Show q^2=prPlease show each step carefully. Thanks.

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### 1 Answer

1. Let's hit this bad boy with the quadratic formula

`x=(-2+-sqrt(4+4(m^2-1)))/2`

The only thing that would make x irrational is that square root. So let's focus on that.

`sqrt(4+4(m^2-1)) = sqrt(4(1+m^2-1))=sqrt(4m^2)=2m`

How nice!

`x=(-2+-2m)/2=-1+-m`

So the roots are not only rational multiples of m, but they are specifically m-1 and -m-1.

2

.`(p^2+q^2)x^2+2q(p+r)x+(q^2+r^2)=0`

Let's use the quadratic formula, eh?

`x = (-2q(p+r)+-sqrt(4q^2(p+r)^2-4(p^2+q^2)(q^2+r^2)))/(2(p^2+q^2))`

We're told the function has real roots. So the value under the square root must be positive (or zero). So...

`4q^2(p+r)^2>=4(p^2+q^2)(q^2+r^2)`

Cancel the 4s and multiply this stuff out

`q^2(p^2+2pr+r^2)>=p^2q^2+p^2r^2+q^4+q^2r^2`

`q^2p^2+2prq^2+q^2r^2>=p^2q^2+p^2r^2+q^4+q^2r^2`

Subtract like quantities from both sides

`2prq^2>=p^2r^2+q^4`

`0>=p^2r^2-2prq^2+q^4`

`0>=(pr-q^2)^2`

A square number cannot be less than zero, so it must be equal to zero.

`0=(pr-q^2)^2`

And I think you can take it from there.