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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).

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  • $\begingroup$ There is no reason to expect $x = 0,1,2$ to correspond in any way to the cases of $0,1,2$ solutions. $\endgroup$ – JHance Jul 21 '14 at 14:47
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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?

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  • $\begingroup$ So when there is one solution and it equals to zero it means there is one x-intercept at the vertex, hence 1 solution and x = 0. When there is a positive number you ± the number, hence there are two x-intercepts (2 solutions). When the number is negative you can't square root the number so there are no x-intercepts/no solutions. $\endgroup$ – Simon Jul 21 '14 at 15:05
  • $\begingroup$ @Simon Yes, that's correct! :) But for the case when there is one solution it is not neccessarily zero, it will be $x = \frac{t}{6}$, whatever $t$ is. $\endgroup$ – naslundx Jul 21 '14 at 15:07
  • $\begingroup$ What I don't understand is how you got this equation... I know the quadratic equation but this seems to look different (is this a modified quadratics equation?). $\endgroup$ – Simon Jul 21 '14 at 15:07
  • $\begingroup$ @Simon I don't know which country you're from and how they teach you to solve quadratics there, but I divide each term by $3$ to get $x^2 + \frac{t}{3}x + \frac{10}{3} = 0$, and then use en.wikipedia.org/wiki/… $\endgroup$ – naslundx Jul 21 '14 at 15:08
  • $\begingroup$ Sorry I meant the quadratics formula: purplemath.com/modules/quads/qform01.gif $\endgroup$ – Simon Jul 21 '14 at 15:10
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$3x^2 + tx + 10 = 0$. 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.

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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 $c=10$.

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