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Most of the other requests for solving simultaneous equations seem to have three equations to use, which my specific problem doesn't have. I also hope to use beginner A-level mathematics, so no calculus, which I've seen in other solutions. The question is:

Find the set of values of k for which the line $y = 2x - k$ meets the curve $y = x^2 + kx - 2$ at two distinct points.

If you could explain any solutions you might have, that would be great.

Edit: thank you all for your the solutions! It has definitely helped. My apologies for saying that calculus was a method of solving, I don’t know the first thing about calculus so thank you for correcting me.

P4dfoot
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4 Answers4

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Setting

\begin{eqnarray} x^2+kx-2&=&2x-k\\ x^2+(k-2)x+(k-2)&=&0\\ \end{eqnarray} The discriminant must be positive in order to obtain two distinct solutions, so

\begin{eqnarray} (k-2)^2-4(k-2)&>&0\\ (k-2)(k-6)&>&0 \end{eqnarray} The graph of this expression with respect to $k$ is a parabola which is concave up, so the expression is positive when $k>6$ and when $k<2$. So the solution set for $k$ is $$(-\infty,2)\cup(6,\infty)$$

John Wayland Bales
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  • @RushabhMehta There is no requirement that $x^2-2x-2=0$ have two solutions. When $k=0$, the graphs of $y=2x$ and $y=x^2-2$ intersect in two distinct points. This is what the question asks. – John Wayland Bales Sep 22 '18 at 19:15
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Hint: Solving the equation $$0=x^2+x(k-2)+k-2$$ we get by the quadratic formula

$$x_{1,2}=-\frac{k-2}{2}\pm\sqrt{\left(\frac{k-2}{2}\right)^2-k+2}$$ Simplifying the discriminant

$$\frac{k^2-4k+4-4k+8}{4}=\frac{k^2-8k+12}{4}$$ Can you finish?

Dr. Sonnhard Graubner
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Well, let's substitute $2x-k$ for $y$ in your quadratic equation, and see what we find;

$$2x-k=x^2+kx-2\to x^2+(k-2)x +(k-2) = 0$$For simplicity, I'll denote $n = k-2$. So, we seek the solutions of the quadratic $$x^2+nx+n=0$$By the quadratic formula, we find that $$x=\frac{-n\pm\sqrt{n^2-4n}}2$$We want to find in which cases there are two solutions to this quadratic, which happens when the discriminant is positive. Hence, we have the inequality $n^2-4n>0$. This implies $$\color{red}{k<2\textrm{ or }k>6}$$

Rushabh Mehta
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The solution range can be determined from the values of k when the line is tangent to the parabola.

$x^2 + kx - 2 = 2x - k\ $ becomes..

$x^2 + (k-2)x -2 + k = 0$

Putting this in perfect square form, $(x + \frac{k-2}{2})^2 = 0$

In which case, $(\frac{k-2}{2})^2 = -2 + k\ $ determines the k values where the line is tangent to the parabola

$(k-2)^2 = -8 + 4k$

$k^2 - 4k + 4 = -8 + 4k$

$k^2 - 8k + 12 = 0$

$(k - 2)(k - 6) = 0$

Therefore $k = 2$ and $k = 6$ are where the line is tangent to the parabola.

Looking at $x^2 + (k-2)x -2 + k = 0$ when $2 < k < 6$, we have all positive terms and hence no roots so the line does not intersect the parabola.

Hence the values of k which intersect the parabola twice are:

$$k<2\ \text{and}\ k>6$$

Phil H
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