Quadratics

Algebra Level 3

$(x-a)(x-b)=p^2$

If $a, b,$ and $p \neq 0$ are real numbers, then what can we say about the nature of roots of the equation above?

real and equal complex cannot be determined real and distinct

1 solution

Arulx Z
Feb 3, 2016

Nature of the roots of quadratic can be determined using the discriminant.

$\left( x-a \right) \left( x-b \right) ={ p }^{ 2 }\\ { x }^{ 2 }-x\left( a+b \right) +\left( ab-{ p }^{ 2 } \right) =0$

Discriminant -

$={ \left( a+b \right) }^{ 2 }-4\left( ab-{ p }^{ 2 } \right) \\ ={ a }^{ 2 }+2ab+{ b }^{ 2 }-4ab+4{ p }^{ 2 }\\ ={ a }^{ 2 }-2ab+{ b }^{ 2 }+4{ p }^{ 2 }\\ ={ \left( a-b \right) }^{ 2 }+4{ p }^{ 2 }$

Since

${ \left( a-b \right) }^{ 2 }+4{ p }^{ 2 } \ge 0$

the roots are real. Since $p\neq 0$ ,

${ \left( a-b \right) }^{ 2 }+4{ p }^{ 2 }>0$

Therefore, the roots are distinct.

Moderator note:

Simple standard approach to find types of roots in a quadatic.

As pointed out by Karthik, the graph visualization yields a much faster approach.

This question can be answered orally if you perform the graph transform from $y=(x-a)(x-b)$ to $y=(x-a)(x-b)-p^{2}$ , and observe the roots ( intersection points with x axis ) in both cases.

Venkata Karthik Bandaru - 5 years, 4 months ago

Nice observation! Didn't think of that before.

Arulx Z - 5 years, 4 months ago

can u explain clearly

Suneel Kumar - 5 years, 3 months ago

You can take a generic graph of equation $y = (x-a)(x-b)$ with two real not necessarily distinct roots.

pb1 pb1

$-p^2$ is always negative. When you do something like $y = (x-a)(x-b)-p^2$ , the graph will always get pushed down.

pb2 pb2

So the parabola will always intersect twice and at distinct places. Hence $y = (x-a)(x-b)-p^2$ always has 2 real distinct roots.

Arulx Z - 5 years, 3 months ago

Quadratics

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