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let : $h: \mathbb{R}\to \mathbb{R}$ ane for any real number

$$h(x^2+x+3)+2h(x^2-3x+5)=6x^2-10x+17$$

then :

$$h(85)=?$$

My Try: $$x=0:h(3)+2h(5)=17\\x=1:h(5)+2h(3)=13\\+\\3h(3)+3h(5)=30\\h(3)+h(5)=10$$

now ?

Martin Sleziak
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user402750
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3 Answers3

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Let $h(x) \equiv Ax+B$, then $$A(x^2+x+3)+B+2A(x^2-3x+5)+2B\equiv 6x^2-10x+17$$

$$ \left \{ \begin{align*} 3A &= 6 \\ -5A &= -10 \\ 13A+2B &= 17 \\ \end{align*} \right.$$

On solving, $(A,B)=\left( 2, -\dfrac{9}{2} \right)$

Alternatively,

  • $x=\dfrac{-1 \pm \sqrt{329}}{2} \implies x^2+x+3=85 \quad \text{and} \quad x^2-3x+5=89 \mp \sqrt{329}$

  • $x=\dfrac{3 \pm \sqrt{329}}{2} \implies x^2-3x+5=85 \quad \text{and} \quad x^2+x+3=89 \pm \sqrt{329}$

  • If you substitute $x=\dfrac{-1+\sqrt{329}}{2}$ and $x=\dfrac{3-\sqrt{329}}{2}$, you can solve $h(85)$ and $h(89-\sqrt{329})$.

  • If you substitute $x=\dfrac{-1-\sqrt{329}}{2}$ and $x=\dfrac{3+\sqrt{329}}{2}$, you can solve $h(85)$ and $h(89+\sqrt{329})$.

Ng Chung Tak
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Denote: $x^2+x+3=t$. Then: $$h(t)+2h(t-4x+2)=6t-16x-1.$$ Plug $x=\frac{1}{2}$ to get: $$h(t)+2h(t)=6t-9 \Rightarrow h(t)=2t-3.$$ Hence: $$h(85)=2\cdot 85-3=167.$$

CryoDrakon
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farruhota
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  • Nice thought, but sadly your solution fails when you plug $x=\frac{1}{2}$ in there, since then you will have that $t=\frac{15}{4}$. – CryoDrakon Jun 14 '18 at 10:05
  • @HeatTheIce, nice objection, yes the variables $x$ and $t$ are interdependent. If they are considered independent from line 2 onward, the argument works. Can you give a counterexample for this method that gives wrong answer? – farruhota Jun 14 '18 at 10:45
  • Everything is fine up until "hence", beacause $t$, or better said $t(x)$ has a fixed value for a fixed value of x. So you only have proven that $h(\frac{15}{4})=\frac{9}{2}$. For you to say that $h(t)=2t-3$ would only be an assumption and not a proof. For it to be proven you need to check it and then verify there are no other solution exists. Sadly I can't give you an example where it wouldn't work, but I think you get the idea why it can't be right. It can be used as a hint for you where the solution is leading you, but a solution on it's own. – CryoDrakon Jun 14 '18 at 11:18
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For a general way of solving this, look at my answer to this question.

When supposing it's a linear function, you also have to prove it can't be anything else, so your solution is complete.

CryoDrakon
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