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

Gold Member
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?

Last edited:

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.

• PeroK
Staff Emeritus
Gold Member
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.

• chwala and PeroK
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
Gold Member
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...

Gold Member
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! • brotherbobby
Homework Helper
Gold Member
2022 Award
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).

• chwala