Relationship between the quadratic being a perfect square and its discriminant?

Question

Am currently doing a question that asks about the relationship between a quadratic and its discriminant.

If we know that the quadratic $ax^2+bx+c$ is a perfect square, then can we say anything about the discriminant?

Specifically, can we be sure that the discriminant equals 0?

So far, I have tried to complete the square for the general quadratic, and got to:

$a((x+\frac{b}{2a})^2-\frac{b^2}{4a^2}+\frac ca)$

But am now stuck. What should I do next, or is there a totally different route I should be taking?

Answer 1

Since the quadratic is a perfect square, this tells you that a root of that quadratic will have multiplicity 2 - you took a linear equation and squared it. If the quadratic vanishes at $x=t$, then so does this linear equation, but we have that linear equation twice by squaring, hence the multiplicity.

So, we can deduce that the discriminant is 0.

Answer 2

By hypothesis, the quadratic is a perfect square if it is something (linear) squared, say $$(ux + v)^2 = u^2x^2 + 2uvx + v^2$$

Then the discriminant $b^2 - 4ac = (2uv)^2 - 4(u^2)(v^2) = 0$

Answer 3

A perfect square takes the form $(px+q)^2$

By expanding the brackets this can be shown to be equal to $p^2x^2+2pqx+q^2$

You want $ax^2+bx+c \equiv p^2x^2+2pqx+q^2$

Comparing coefficients of $x^2$ gives you $a=p^2$

Comparing coefficients of $x$ gives you $b=2pq$

Comparing constant terms gives you $c=q^2$

The discriminant is $b^2-4ac$

Using the expressions above means that $b^2-4ac=(2pq)^2-4(p^2)(q^2)=4p^2q^2-4p^2q^2=0$

Answer 4

The quadratic has two equal roots like d if it is perfect square and we may write:

$(x-d)^2=x^2-2dx +d^2$

Then:

$\Delta=4d^2-4d^2=0$

That is the discriminant is zero when the quadratic is perfect square.

Answer 5

Let us assume that $ax^2+bx+c=d=(jx+k)^2 $

\begin{equation} d=(jx+k)^2\implies j^2x^2+2jkx+k^2-d=0\implies a=j^2\quad b=2jk\quad c=k^2-d\\ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} =\frac{-2jk\pm\sqrt{2^2j^2k^2-4j^2(k^2-d)}}{2j} =\frac{-2jk\pm 2j\sqrt{d}}{2j} =\pm\sqrt{d}-k \end{equation}

By definition above, d is a perfect square and the discriminant (just before and after the last equal sign) supports that. If $ax^2+bx+c=0$, then the discriminant is zero.