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If the roots of $9x^2-2x+7=0$ are $2$ more than the roots of $ax^2+bx+c=0$, then value of $4a-2b+c=$?

My approach is: Let the roots of $ax^2+bx+c=0$ are $\alpha$ and $\beta$. So, $$\alpha+\beta=-\frac{b}{a}\\ \alpha\beta=\frac{c}{a}$$ Therefore the roots of $9x^2-2x+7=0$ are $(\alpha+2)$ and $(\beta+2)$. So, $$\alpha+\beta+4=\frac{2}{9}\implies4-\frac{b}{a}=\frac{2}{9}\implies\frac{b}{a}=\frac{34}{9}$$ and \begin{align*} (\alpha+2)(\beta+2)=\frac{7}{9}\\ \Rightarrow\alpha\beta+2(\alpha+\beta)+4=\frac{7}{9}\\ \Rightarrow\frac{c}{a}-2\frac{b}{a}+4=\frac{7}{9}\\ \Rightarrow\frac{4a-2b+c}{a}=\frac{7}{9}\\ \end{align*} So, my final equations are: $\dfrac{b}{a}=\dfrac{34}{9}$ and $\dfrac{4a-2b+c}{a}=\dfrac{7}{9}$. My mind says I'm pretty close to the solution but I can't find it out.

There is another similar question: If the roots of $px^2+qx+r=0$ are $2$ more than the roots of $ax^2+bx+c=0$, then what will be the expression of $r$ in terms of $a$, $b$, and $c$?

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  • $\begingroup$ The roots are not real though :-) $\endgroup$ – Math-fun Mar 30 '17 at 9:19
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Hint: Find the roots of equation $9x^2-2x+7=0$ by solving this let say the roots be $\alpha$ and $\beta$ (You know the value of $\alpha$ and $\beta$.) then the roots of other equation must be $\alpha-2$ and $\beta-2$ form a equation using $$(x-(\alpha-2))(x-(\beta-2))$$ compare the coefficient. then put the value in $4a-2b+c.$

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My way:

If $p$ is a root of $9x^2-2x+7=0, p-2$ will be a root of $ax^2+bx+c=0$

Now writing $p-2=q\iff p=q+2$ in $9x^2-2x+7=0,$

we get $0=9(q+2)^2-2(q+2)+7=9q^2+34q+39$

So, we need $$\dfrac a9=\dfrac b{34}=\dfrac c{39}\ \ \ \ (1)$$

So, with the given conditions, $a,b,c$ can assume any set of non-zero finite values that honor $(1)$

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  • $\begingroup$ @kingW3, I was searching for the mistake. Thanks $\endgroup$ – lab bhattacharjee Mar 30 '17 at 9:16
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You can check the other answers for (possible faster) alternatives, but you were doing fine.

So, my final equations are: $\dfrac{b}{a}=\dfrac{34}{9}$ and $\dfrac{4a-2b+c}{a}=\dfrac{7}{9}$. My mind says I'm pretty close to the solution but I can't find it out.

Notice that the system $$\left\{ \begin{array}{l}\dfrac{b}{a}=\dfrac{34}{9} \\[8pt] \dfrac{4a-2b+c}{a}=\dfrac{7}{9} \end{array}\right. \iff \left\{ \begin{array}{l}34a-9b=0 \\[8pt] 29 a - 18 b + 9 c = 0 \end{array}\right.$$ has an infinite number of solutions, given by: $$\left\{ \begin{array}{l} a = 9t \\ b = 34t \\ c = 39t \end{array}\right. \quad t \in \mathbb{R_0}$$ This is logical because we can simply divide $ax^2+bx+c=0$ (for $a$ non-zero) by $a$ to obtain a quadratic equation with the same roots.

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Since $ax^2+bx+c=0$ has the same roots as $x^2+\frac{b}{a}x+\frac{c}{a}=0$ we can notice that if $(a_1,b_1,c_1)$ satisfies the requirements then also does $(ta_1,tb_1,tb_2)$ since $4a-2b+c\neq 0$ we can get $4a-2b+c=\frac{7t}{9}$ for any $t$.

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We find the roots of $9x^2-2x+7=0$ using the quadratic formula:

\begin{align}x&=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ &=\frac{-(-2)\pm\sqrt{(-2)^2-4\times9\times7}}{2\times 9}\\ &=\frac{2\pm\sqrt{4-252}}{18}\\ &=\frac{2\pm\sqrt{-248}}{18}\\ &=\frac{2\pm2i\sqrt{62}}{18}\\ &=\frac{1\pm i\sqrt{62}}{9}\end{align}

Or through completing the square$^*$:

\begin{align}9x^2-2x+7&=0\\ x^2-\frac 29x+\frac 79&=0\\ \left(x-\frac19\right)^2-\left(\frac 19\right)^2+\frac 79&=0\\ \left(x-\frac 19\right)^2&=\frac 1{81}-\frac 79\\ \left(x-\frac 19\right)^2&=-\frac{62}{81}\\ x-\frac 19&=\pm \sqrt{-\frac{62}{81}}\\ x-\frac 19&=\pm\frac{i\sqrt{62}}{9}\\ x&=\frac 19 \pm \frac{i\sqrt{62}}{9}\\ &=\frac{1\pm i\sqrt{62}}{9}\end{align}

We can then say that the roots of $ax^2+bx+c=0$ are $$\frac{1\pm i\sqrt{62}}{9}-2$$

So, we can say \begin{align}ax^2+bx+c&=\left(x-\left(\frac{1+ i\sqrt{62}}{9}-2\right)\right)\left(x-\left(\frac{1 - i\sqrt{62}}{9}-2\right)\right)\\ &=x^2-x\left(\frac{1 - i\sqrt{62}}{9}-2\right)-x\left(\frac{1 + i\sqrt{62}}{9}-2\right)+\left(\frac{1 - i\sqrt{62}}{9}-2\right)\left(\frac{1 + i\sqrt{62}}{9}-2\right)\\ &=x^2-\left(\frac{34}{9}\right)x+\frac {13}{3}\end{align}

So we can say that $a=1$, $b=-\dfrac{34}{9}$, and $c=\dfrac{13}{3}$ and therefore \begin{align}4a+2b+c&=4\times 1+2\times-\frac{34}{9}+\frac{13}{3} \\ &= 4-\frac {68}{9}+\frac{13}{3}\\ &=\frac 79\end{align}


$^*$Completing the square:

For $ax^2+bx+c$, we can say that $$x^2+\frac ba x + \frac ca = \left(x-\frac {b/a}{2}\right)^2-\left(\frac {b/a}{2}\right)^2-\frac ca$$



For the general case:

The roots of $px^2+qx+r=0$ are $$x=\frac{-q\pm \sqrt{q^2-4pr}}{2p}$$

We can also say that the roots of $ax^2+bx+c$ are

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=2-\frac{-q\pm \sqrt{q^2-4pr}}{2p}$$

We can then solve this for $r$:

\begin{align}\frac{-b\pm\sqrt{b^2-4ac}}{2a}&=2-\frac{-q\pm \sqrt{q^2-4pr}}{2p}\\ 2-\frac{-b\pm\sqrt{b^2-4ac}}{2a}&=\frac{-q\pm \sqrt{q^2-4pr}}{2p}\\ 4p-2p\frac{-b\pm\sqrt{b^2-4ac}}{2a}&=-q\pm \sqrt{q^2-4pr}\\ 4p+q-2p\frac{-b\pm\sqrt{b^2-4ac}}{2a}&=\pm \sqrt{q^2-4pr}\\ \left(4p+q-2p\frac{-b\pm\sqrt{b^2-4ac}}{2a}\right)^2&=q^2-4pr\\ q^2-\left(4p+q-2p\frac{-b\pm\sqrt{b^2-4ac}}{2a}\right)^2&=4pr\\ \frac{q^2-\left(4p+q-2p\frac{-b\pm\sqrt{b^2-4ac}}{2a}\right)^2}{4p}&=r\\ \frac{-8a^2(2p+q)+2a(4p+q)\left(-b\pm\sqrt{b^2-4ac}\right)-p\left(-b\pm\sqrt{b^2-4ac}\right)^2}{4a^2}&=r\end{align}

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