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I am given an $x$-intercept of $-3-\sqrt{7}$ and I am asked to find the other intercept. I am having trouble since I don't have any other information but the given $x$-intercept. My guess is that the other $x$-intercept is $-3+\sqrt{7}$ but I don't know how to show it mathematically if it's right. Or rather can I have any other intercepts since I can make my own vertex and axis of symmetry. There is no restriction at all. Please help.

Here is the exact problem. One zero of a quadratic function (whose graph is a parabola) is $-3-\sqrt{7}$. What is the other zero?

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    $\begingroup$ In the absence of any other information, any point (including this point) can be the other $x$-intercept. $\endgroup$ – GoodDeeds Feb 1 '16 at 13:29
  • $\begingroup$ You can have infinitely many such parabolas. $\endgroup$ – DeepSea Feb 1 '16 at 13:33
  • $\begingroup$ Could you post the problem in its entirety? One suspects that there is more context given. $\endgroup$ – Cameron Buie Feb 1 '16 at 13:35
  • $\begingroup$ just posted it. I really think I can choose any other x-intercept I want as long as it's a real number. $\endgroup$ – marg_ocruz Feb 1 '16 at 13:40
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You have to have more information than that. You can construct infinite number of parabolas with arbitrary x-axis intersects (zero, one or two).

For example you can write $y = (x-x_0)(x-x_1)$ toget a parabola that intersects at $x_1$ and $x_2$, but there are more ways than than. You could for example take $x = y^2+3+\sqrt7$ which is also a parabola - just that it doesn't happen to be facing upwards or downwards.

If you on the other hand requires the parabola to be expressable as $y = P(x)$ where $P$ is an polynomial with rational coefficioents then the solution is unique. The polinomial is uniquely determined up to rational scaling.

Suppose that the coefficient of the $x^2$ term is one, then we have

$$P(x) = (x+3+\sqrt7)(x-c) = x^2 + (3 + \sqrt7-c)x+(3+\sqrt7)cx$$

Now we require $3+\sqrt7-c$ to be rational which makes $c = q+\sqrt7$ for some rational $q$, but we also require $(3+\sqrt7)c = (3+\sqrt7)(q+\sqrt7) = 3q + (3+q)\sqrt7 + 49$ to be rational. But since $\sqrt7$ is irrational we must have that $3+q=0$ and consequently that $c = q+\sqrt7 = -3+\sqrt 7$.

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Hint: Assume the parabola is the graph of the equation $$y=a(x-b)(x-c)$$ You know one of $b$ and $c$, say $c=-3-\sqrt{7}$ You are free to choose $a$ and $b$.

This means that the other intercept, and the vertical scale factor, can be chosen at will. You could even have the two intercepts coincide, so that the parabola just touches the $x$-axis at $(-3-\sqrt{7},0)$.

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