Given the equation of the parabola, for what values of $t$ will the quadratics have $0$, $1$, or $2$ solutions?

Question

Given the equation of the parabola, for what values of $t$ will the quadratics have $0$, $1$, or $2$ solutions? Equation: $$0 = 3x^2 + tx + 10$$

Can you please explain the answer in simple terms, step by step if you can. How do I figure this out? I tried plugging in the $0$, $1$, and $2$ for the $x$ variables and solve for $t$ but that only gave me a correct solution for $0$ (when I plugged in $0$ $t$ had $0$ $x$ intercepts, but when I plugged in $1$ and later $2$, both had $2$ solutions so there must be something wrong there).

Answer 1

Hint

The solution for your equation is $$x = -\frac{t}{6} \pm \sqrt{\frac{t^2}{36} - \frac{10}{3}}$$

The stuff below the square root sign ($\frac{t^2}{36} - \frac{10}{3}$) is the important thing. You get exactly one solution when it equals zero, two solutions when it is positive and no (real) solutions when it is negative.

Can you find out when the different cases occur? And why this is so?

Answer 2

$3x^2 + tx + 10 = 0$. Discriminant: $\Delta = b^2 - 4ac = t^2 - 120$.

If $\Delta > 0$ two real solutions, that is, $t^2 - 120 > 0 \ \Rightarrow \ t < -2\sqrt{30}$ and $t > 2\sqrt{30}$

If $\Delta = 0$ one real solution, that is, $t^2 - 120 = 0 \ \Rightarrow \ t =\pm 2\sqrt{30}$

If $\Delta < 0$ no real solutions.

Answer 3

In $\mathbb{R}$ an equation of the form: $$z^{2}=c$$ has $2$ solutions if $c>0$ (then $z=\pm\sqrt{c}$), has $1$ solution if $c=0$ (then $z=0$) and has no solutions if $c<0$.

If $a\neq0$ then the following statements are equivalent:

$$ax^{2}+bx+c=0$$

$$\left(x+\frac{b}{2a}\right)^{2}=\frac{b^{2}-4ac}{4a^{2}}$$

Note that the last line can be written as $z^{2}=c$ for $z=x+\frac{b}{2a}$ and $c=\frac{b^{2}-4ac}{4a^{2}}$. This leads to:

If $\frac{b^{2}-4ac}{4a^{2}}>0$ or equivalently $b^{2}-4ac>0$ then there are $2$ solutions.

If $\frac{b^{2}-4ac}{4a^{2}}=0$ or equivalently $b^{2}-4ac=0$ then there is $1$ solution.

If $\frac{b^{2}-4ac}{4a^{2}}<0$ or equivalently $b^{2}-4ac<0$ then there are $0$ solutions.

So crucial here is the sign of $D:=b^{2}-4ac$ (the so-called discriminant).

Apply this on your question where $a=3$, $b=t$ and $c=10$.