Find $h(85)$ if $h(x^2+x+3)+2h(x^2-3x+5)=6x^2-10x+17$

Question

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 ?

Answer 1

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

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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 $h(89+\sqrt{329})$.