Show that the equation has real and distinct roots

**PythagorasNeophyte** · February 9th 2011, 11:12 PM

Question: show that the roots of the equation $(px-1)^2 +3px -5 = 0$ are real and distinct for all real values of $p$ and $p \neq 0$

My workings:
$(px-1)^2 +3px -5 = 0$

$p^2x^2 -2px +1 +3px -5 = 0$

$p^2x^2 +px -4= 0$

Here $a = p^2, b = p, c = -4$ . Now

$b^2 -4ac = p^2 -4p^2(-4)$

$= p^2 + 16p^2$

$= 17p^2$

This doesn't prove anything. Can you please help me? I don't have anyone to guide me.

**earboth** · February 9th 2011, 11:18 PM

Originally Posted by PythagorasNeophyte

Question: show that the roots of the equation $(px-1)^2 +3px -5 = 0$ are real and distinct for all real values of $p$ and $p \neq 0$

My workings:
$(px-1)^2 +3px -5 = 0$

$p^2x^2 -2px +1 +3px -5 = 0$

$p^2x^2 +px -4= 0$

Here $a = p^2, b = p, c = -4$ . Now

$b^2 -4ac = p^2 -4p^2(-4)$

$= p^2 + 16p^2$

$= 17p^2$

This doesn't prove anything. Can you please help me? I don't have anyone to guide me.

A square is allways positive or zero. If you multiply a square by a positive number the result is positive too.

You get two different values if the discriminant unequals zero. Therefore you'll get real and distinct roots of the equation if $17p^2\ne 0 ~\implies~p\ne 0$

**CaptainBlack** · February 10th 2011, 08:22 PM

Originally Posted by PythagorasNeophyte

Question: show that the roots of the equation $(px-1)^2 +3px -5 = 0$ are real and distinct for all real values of $p$ and $p \neq 0$

My workings:
$(px-1)^2 +3px -5 = 0$

$p^2x^2 -2px +1 +3px -5 = 0$

$p^2x^2 +px -4= 0$

Suppose $$$p>0$ , then by Descartes rule of signs there is exactly one positive roots, and exactly one negative root.

Suppose $$$p<0$ , then by Descartes rule of signs there is exactly one positive root and one negative root.

QED

CB

**HallsofIvy** · February 11th 2011, 03:39 AM

Originally Posted by PythagorasNeophyte

Question: show that the roots of the equation $(px-1)^2 +3px -5 = 0$ are real and distinct for all real values of $p$ and $p \neq 0$

My workings:
$(px-1)^2 +3px -5 = 0$

$p^2x^2 -2px +1 +3px -5 = 0$

$p^2x^2 +px -4= 0$

Here $a = p^2, b = p, c = -4$ . Now

$b^2 -4ac = p^2 -4p^2(-4)$

$= p^2 + 16p^2$

$= 17p^2$

This doesn't prove anything. Can you please help me? I don't have anyone to guide me.

Do you recall why you calculated that? It is the discriminant of the equation and a quadratic equation has distinct real roots if and only if its discriminant is positive. Since $p\ne 0$ is a condition of the problem, what can you say about $p^2$ ?

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