Difficult quadratics question

Question

$$\frac{x^3}{2}-kx^2+4kx-32=0$$
Find $k$ such that there are 2 solutions

I don't understand the working here. Specifically for when they consider -kx^2 +4kx = 0. I don't get how solving x from this allows us to determine what values will give us 2 solutions?

Can you instead factor the cubic in to a linear and quadratic term. Then use the discrimant for the quadratic to solve when it has 1 solution?

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Answer 1

${x^3\over2}-kx^2+4kx-32=(1/2)(x^3-64-2kx^2+8kx)=(1/2)(x-4)(x^2+4x+16-2kx)$

To have two roots is to have a repeated root, so either the polynomial $x^2+(4-2k)x+16$ is a perfect square, or else it has $x=4$ as a root. So....

Answer 2

Hint

$$\frac{x^3}{2}-kx^2+4kx-32=0$$ is a cubic equation.

In order its has two real roots, it must three real roots two of them being identical.

If you look at the Wikipedia page, this means that $$\Delta=18 a b c d - 4 b^3 d + b^2 c^2 - 4 a c^3 - 27 a^2 d^2=0$$ Using $a=\frac 12$, $b=-k$, $c=4k$, $d=-32$, this leads to $$\Delta=16 k^4-256 k^3+1152 k^2-6912=16(k^4-16 k^3+72 k^2-432)=0$$ By inspection $k=-2$ is a root. Then factoring $$\Delta=16(k+2)(k^3-18 k^2+108 k-216)=0$$ But, you can recognize that the last term is just $(k-6)^3$. So $$\Delta=16(k+2)(k-6)^3$$

I am sure that you can take it from here.

Answer 3

Let $$f(x)=\frac{x^3}{2}-k x^2+4 k x-32$$

Note that $f(4)=0$ so $$f(x)=\frac{1}{2}(x-4)\left(x^2-(2k-4)x+16\right)$$

If $4$ is a root of the quadratic term, then the other root is $4$ as well since the product of the roots is $16$. In this case $f$ has three equal roots.

Otherwise, $-4$ must be a double root of $x^2-(2k-4)x+16$

Therefore

$$(-4)^2-(2k-4)(-4)+16=8k+16=0$$

and $k=-2$