Find the range of values of k such that for all $x$, $kx−2 \le x^2$
I found the discriminant of the equation which I thought would be greater than equal to zero, so that there will be at least one real root of the equation.
However, the solution states that the discriminant should be less than equal to zero. Why is the answer less than?
The discriminant is $k^2-8$
Thanks:)
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$$kx-2\le x^2\iff x^2-kx+2\ge0\iff \left(x-\frac k2\right)^2+2-\frac{k^2}{4}\ge 0$$ $\because \left(x-\frac k4\right)^2\ge 0\quad \forall \ \ x, k \in \mathbb R$ $$ \therefore 2-\frac{k^2}{4}\ge 0\iff k^2\le 8\iff |k|\le 2\sqrt2$$ $$k\in [-2\sqrt2, 2\sqrt2]$$
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In general, the quadratic polynomial: $ax^2+bx+c$ will be non-negative iff the upward parabola is just touching or above the X-axis i.e. $$ax^2+bx+c\ge 0 \iff a>0, \ b^2-4ac\le 0$$
Similarly, the quadratic polynomial: $ax^2+bx+c$ will be non-positive iff the downward parabola is just touching or below the X-axis i.e. $$ax^2+bx+c\le 0 \iff a<0, \ b^2-4ac\le 0$$
As per given question: $$x^2-kx+2\ge 0\iff a=1>0, \ \ (-k)^2-4(1)(2)\le 0$$ $$k^2-8\le 0\iff |k|\le 2\sqrt2$$
We need $$x^2-kx+2\ge 0$$ The discriminant must indeed be non-positive. Why? Let’s assume $D\gt 0$. This would mean that our quadratic equation has two roots, i.e. it cuts the $x-$ axis at two distinct points. Notice that the leading coefficient of $x^2$ is $+1$, and the parabola is upward-facing. The existence of two roots means that the quadratic will assume negative values between the two roots, and hence violate the inequality. For example, see this picture below:
If instead the quadratic has atmost one root, then it can never ‘dip’ below the $x$-axis and will always be positive. So, the required condition is $D\le 0$.