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$$\frac{x^3}{2}-kx^2+4kx-32=0$$
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?

Edit: I can only post links to my image because I need 10+ reputation to do so. Upvote this and I'll edit it when I have enough. Cheers.

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    `Upvote this and I'll edit it when I have enough` How about *you* transcribe your question in [MathJax](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference) then I'll upvote it. – 2017-02-25
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    It's not clear from the image as to what you're asking. If you don't know how to use MathJax, at least write the problem clearly in words and text-based math notation. – 2017-02-25
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    You can factor $x-4$, but I don't think that the hint regarding the quadratic can really help. – 2017-02-25

3 Answers 3

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

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....

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    Thanks. When x = 4 -> k = -6. Why is this invalid? Doesn't tell you that (x-4) is repeated, but doesn't tell you anything about the last factor? – 2017-02-25
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    When $x=4$ is a root of the quadratic, then $k=6$ (not $-6$), the quadratic is $x^2-8x+16=(x-4)^2$, so as Lozenges notes $x=4$ is a *triple* root of the original cubic, which thus has just one root, not two. – 2017-02-25
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    Ah right, cheers. – 2017-02-25
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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.

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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 $k=-2$