Show algebraically that the graph of $y=x^2 + kx - 2$ will cut the $x$-axis twice for all values of $k$

Question

A quadratics question.

Show algebraically that the graph of $y=x^2 + kx - 2$ will cut the $x$-axis twice for all values of $k.$

I recently asked a similar question, but this problem seems broader in scope.

Answer 1

Use the quadratic function, with a= 1, b= k, and c= -2.

There's three different outcomes for a quadratic:

${b^2-4ac}\gt 0 \rightarrow$ we get two solutions, which means we have two x-intercepts.

${b^2-4ac}=0 \rightarrow$ we get one solution, which means we have one x-intercept

${b^2-4ac}\lt 0 \rightarrow$ no real solutions, which means no x-intercepts. You can think of the parabola of the axis being above the x-axis if it opens upwards, or below the x-axis if it opens downwards.

Answer 2

We are basically attempting to show that

$$ y = x^2 +kx - 2$$

Equals 0, for exactly two, distinct x, for any choice of real number k. A trick here is to basically factor the expression, and show that the factors give rise to distinct roots.

So we want to find an $a,b$ such that

$$ (x-a)(x-b) = x^2 + kx - 2$$

And show that regardless of the $k$, $a \ne b$.

This amounts to solving

$$ x^2 + kx - 2 = 0 \rightarrow x = \frac{-k \pm \sqrt{k^2+8}}{2}$$

And from here we let $a = \frac{-k + \sqrt{k^2 + 8}}{2}$, $b = \frac{-k - \sqrt{k^2 + 8}}{2}$.

now what is left to be shown is that, for ANY choice of k

$$\frac{-k + \sqrt{k^2 + 8}}{2} \ne \frac{-k - \sqrt{k^2 + 8}}{2} $$

Of course that is simply checking that

$$ -k + \sqrt{k^2 + 8}\ne -k - \sqrt{k^2 + 8}$$

And more importantly that:

$$ \sqrt{k^2 + 8}\ne- \sqrt{k^2 + 8} $$

Observe that the only time these are equal is if $k^2 = -8 \rightarrow k = \pm i 2\sqrt{2}$, but we already restricted our attention to $k$ being a REAL number, so we then can safely conclude that these two expression are never EQUAL.

and therefore, this quadratic ALWAYS has distinct roots.

And thus, it touches the x-axis exactly two times regardless of $k$.

---

@Joanpemo, points out rightly,

that while we have showing if $k$ is REAL that the roots are distinct, what have yet to show is that the ROOTs themselves are REAL too (otherwise the question of intersecting with the x-axis is pointless to begin with).

So to show

$$ \frac{-k \pm \sqrt{k^2+8}}{2} $$

is real, we need to show

$$ -k \pm \sqrt{k^2+8} $$

is real, which reduces to showing that

$$ \sqrt{k^2 + 8}$$

is real, this follows from the fact that $k^2 > 0$ since $k$ is real, and thus this expression is always the square root of a positive number (which is well defined). So we conclude the roots themselves do lie on the real x-axis.

Answer 3

for graph Y = x^2 + Kx -2 to cut x - axis y coordinate must be zero thus,

x^2 + Kx -2 = 0

Now for this equation for solution to be finitly 2 , b^2 - 4ac > 0.

here b = K , a = 1 and c = -2.

for every value of K , b^2 - 4ac will be greater than zero.

for example k = 0 => (0)^2 - 4(1)(-2) = 8

for k = -1 => (-1)^2 - 4(1)(-2) = 9

hence for all value of K it will be greater than zero so now we can find the solution of an equation.