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

• chwala
This is equivalent to the discriminant condition D≥0, which is satisfied when q^2-4pd≥0. Therefore, the statement is proven.
chwala
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:
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
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
Office_Shredder said:
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
brotherbobby said:
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...

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
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

## 1. What is a quadratic equation?

A quadratic equation is an algebraic equation in the form of ax^2 + bx + c = 0, where a, b, and c are constants and x is the variable. It is called a quadratic equation because the highest power of the variable is 2.

## 2. How do you determine if a quadratic equation has real roots?

A quadratic equation has real roots if the discriminant, b^2 - 4ac, is greater than or equal to 0. If the discriminant is equal to 0, the equation has one real root, and if it is greater than 0, the equation has two real roots.

## 3. What is the significance of the equation (x-a)(x-b)=b^2?

This equation represents a quadratic equation with the roots a and b. The product of the roots is equal to the constant term, b^2. This can be used to solve for the roots of the equation.

## 4. How do you solve the equation (x-a)(x-b)=b^2 to find the roots?

To solve this equation, we can use the zero product property, which states that if the product of two factors is equal to 0, then at least one of the factors must be equal to 0. So, we can set each factor equal to 0 and solve for x, giving us the roots a and b.

## 5. How can you show that the equation (x-a)(x-b)=b^2 has real roots?

To show that this equation has real roots, we can use the discriminant, b^2 - 4ac, where a=1, b=-(a+b), and c=b^2. Substituting these values into the discriminant, we get (a+b)^2 - 4(1)(b^2) = a^2 + 2ab + b^2 - 4b^2 = a^2 - 2b^2. Since a^2 and b^2 are always positive, the discriminant is always greater than or equal to 0, meaning the equation has real roots.

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