What must the discriminant be?

Algebra Level 3

$(p^{2} + q^{2})x^{2} - 2prx - q^{2} + r^{2} = 0$

In what correlation do the parameters $p, q$ and $r$ need to be, so that the above equation has distinct real solutions?

r^2 < p^{2}+q^{2}

r^{2}+p^{2} > q^2

r^{2}+q^{2} > p^2

q^{2}+r^{2} = p^2

2 solutions

Ismet Cosic
Jul 2, 2016

$a= p^{2}+q^{2}$

$b=-2pr$

$c= -q^{2}+r^{2}$

$D ≥ 0$

$b^{2}-4ac ≥ 0$

$4q^{2}(p^{2} + q^{2} -r^{2}) ≥ 0$

$p^{2} + q^{2} ≥ r^{2}$

$\boxed{r^{2} ≤ p^{2} + q^{2}}$

No... For distinct solutions Discriminant>0 and not $\ge 0$ so there must be a strict inequality.

Rishabh Jain - 4 years, 11 months ago

my bad. sorry.

ismet cosic - 4 years, 11 months ago

No problem... :-)

Rishabh Jain - 4 years, 11 months ago

Note that you cannot divide by $4q^2$ which could be 0.

For example, we will still have real roots if $q = 0$ , $r = 1 , p = 0$ , which doesn't satisfy your conditions.

I have edited the problem for clarity.

Calvin Lin Staff - 4 years, 11 months ago

J D
Jul 3, 2016

To have real SOLUTIONS (plural!) the discriminant should technically be greater than zero (and not equal to it). If it were equal to zero, then there is only one real solution.

Even if the discriminant is equal to zero, it has two solutions. It just happens that they're the same. Two identical solutions if you wish.

ismet cosic - 4 years, 11 months ago

What must the discriminant be?

2 solutions

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