Show that ##(x-a)(x-b)=b^2## has real roots- Quadratics

[chwala]

Homework Statement

Show that $(x-a)(x-b)=b^2$ has real roots

Relevant Equations

discriminant

If we have a quadratic equation, $px^2+qx+d$ then it follows that for real roots; The discriminant
$D= q^2-4pd≥0$ therefore on expanding $(x-a)(x-b)=b^2$ we get,
$x^2-bx-ax+ab-b^2=0$
$a^2+2ab+b^2-4ab+4b^2≥0$
$a^2-2ab+b^2+4b^2≥0$,
$(a-b)^2+4b^2≥0$
since, $(a-b)^2 ≥0$ and $4b^2≥0$ is true for any value of $a,b ∈ℝ$ then our proof is complete. Thanks guys Bingo!
Is there another way of proving this?

 

[brotherbobby]

First, please differentiate between the coefficients $A, B, C$ of the general equation $Ax^2+Bx+C=0$ and those ($a,b,c$) of yours.
Rearrange your equation and apply the real condition for the discriminant and see where that leads you.

 

[Office_Shredder]

Another approach. There's an obvious value of $x$ such that $y=(x-a)(x-b)-b^2$ has a point lying below the x axis. It also has a positive $x^2,$ coefficient, so just intersect the x-axis in two places.

For your method, do you know how to actually prove the last line is true? It's not a super tricky inequality, but I think requires justification.

 

[brotherbobby]

Another approach. There's an obvious value of $x$ such that $y=(x-a)(x-b)-b^2$ has a point lying below the x axis. It also has a positive $x^2,$ coefficient, so just intersect the x-axis in two places.

For your method, do you know how to actually prove the last line is true? It's not a super tricky inequality, but I think requires justification.

Yes. I don't want to give it away for the OP @chwala to see how it is, but the discriminant ($\mathscr{D} = B^2-4AC$) came as a sum of two squares in the end and therefore has to be greater than or equal to 0.

 

[chwala]

First, please differentiate between the coefficients $A, B, C$ of the general equation $Ax^2+Bx+C=0$ and those ($a,b,c$) of yours.
Rearrange your equation and apply the real condition for the discriminant and see where that leads you.

I will look into this later...

 

[chwala]

If we have a quadratic equation, $px^2+qx+d$ then it follows that for real roots; The discriminant
$D= q^2-4pd≥0$ therefore on expanding $(x-a)(x-b)=b^2$ we get,
$x^2-bx-ax+ab-b^2=0$
$a^2+2ab+b^2-4ab+4b^2≥0$
$a^2-2ab+b^2+4b^2≥0$,
$(a-b)^2+4b^2≥0$
since, $(a-b)^2 ≥0$ and $4b^2≥0$ is true for any value of $a,b ∈ℝ$ then our proof is complete. Thanks guys Bingo!

 

[Steve4Physics]

There is a simple ‘geometrical’ view which shows the proposition is obviously true. It may be equivalent to what @Office_Shredder said in Post #3.

y = (x-a)(x-b) is an ‘upright’ parabola with real roots (x=a and x=b). So its vertex is always below the x-axis.

Shifting the parabola in the downwards (-y direction) by an amount S (S≥0) gives a new parabola, y = (x-a)(x-b) – S.

This new parabola must also have real roots because the (downwards-shifted, upright) parabola must still intersect the x-axis at two points.

That means (x-a)(x-b) = S always has real roots (provided S≥0).