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Find the equation whose roots are each six more than the roots of $x^2 + 8x - 1 = 0$

I must use Vieta's formulas in my solution since that is the lesson we are covering with our teacher.

My solution:

Let p and q be the roots of the quadratic.

$$\begin{align} p + q = & -8 \\ pq = & -1 \end{align}$$

If the roots are each six more than the roots of the quadratic, then we will have:

$$\begin{align} p + q + 12 = & -8 \\ p + q = & -20 \\ pq = & - 3 \end{align}$$

Also,

$$ \begin{align} (p+6)(q+6) = & pq + 6p + 6q + 36\\ =&pq + 6(p+q) + 36 \\ =&-3 + 6(-8) + 36 \\ =&-3 - 48 + 36 \\ &36 - 51 = -15 \end{align} $$

Hence, -15 = constant.

Thus, the quadratic equation is $x^2 + 20x - 15$

My worksheet gives an answer of $x^2 - 4x - 13$, how am I wrong?

Thanks!

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  • $\begingroup$ The old roots add up to $-8$. So the new roots add up to $-8+6+6=4$, and herefore the new coefficient of $x$ is $-4$. $\endgroup$ – André Nicolas May 1 '15 at 0:39
  • $\begingroup$ As for their product, where did you take $pq=-3$? $\endgroup$ – Bernard May 1 '15 at 0:40
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    $\begingroup$ I understand that you said you must use Vieta's formula but the answer is as simple as $(x-6)^2 +8(x-6) -1 =0$ which quickly simplifies to the required answer. If you encounter this sort of problem in future and aren't constrained to use Vieta's Formula, this is a more direct route. $\endgroup$ – Deepak May 1 '15 at 1:11
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Let $r=p+6$ and $s=q+6$ be the roots of the quadratic you're trying to find. That quadratic's coefficients are $-(r+s)$ and $rs$. But

$$\begin{align} r+s&=(p+6)+(q+6)\\ &=p+q+12\\ &=-8+12\\ &=4 \end{align}$$ and $$\begin{align} rs&=(p+6)(q+6)\\ &=pq+6(p+q)+36\\ &=-1+6(-8)+36\\ &=-1-48+36\\ &=-13 \end{align}$$

so the quadratic you're looking for is

$$x^2-(r+s)x+rs=x^2-4x-13$$

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Let $x_1,x_2$ be the roots of the equation: $x^2+8x-1 = 0$, and let $y_1 = x_1 +6, y_2 = x_2+6 \Rightarrow y_1+y_2 = x_1+x_2+12=-8+12 = 4, y_1y_2 = (x_1+6)(x_2+6)=x_1x_2+6(x_1+x_2)+36=-1+6\cdot (-8) + 36=-13\Rightarrow y^2-4y-13=0$ is the sought after equation.

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The original quadratic equation is given to you as $x^2 + 8x - 1 = 0$ . Let it's roots be $(x_1 , x_2)$ . The roots of required equation be $(\alpha ,\beta)$ (say).

Now, $$x_1 + x_2 = -8 \\ x_1 x_2 = -1$$

And $$\alpha = x_1 + 6 \\ \beta = x_2 + 6 $$

So, $$\alpha + \beta = x_1 + x_2 + 12 \\ \alpha \beta = (x_1 + 6)(x_2 + 6) = x_1 x_2 + 6(x_1 + x_2) + 36 $$

You already know the values of $\color{red}{x_1 x_2 = -1} ; \ \color{blue}{x_1 + x_2 = -8}$ . So, just plug-in the values in the above formula.

$$\begin{align} &\alpha + \beta = \color{blue}{x_1 + x_2} + 12 = \color{blue}{-8} + 12 = 4 \\ &\alpha \times \beta = \color{red}{x_1 x_2} + 6\color{Blue}{(x_1 + x_2)} + 36 = \color{red}{-1} + 6\color{blue}{(-8)} + 36 = -13\end{align}$$

So, you can just finalize it and write the quadratic equation as : $$\boxed{\color{brown}{x^2 - 4x - 13}}$$

NOTE : Any quadratic equation can be written as : $x^2 - \color{blue}{S} x + \color{red}{P} $ .

Where: $\color{blue}{S}$ = Sum of Zeroes(or roots) of the equation.

$\color{red}{P}$ = Product of Zeroes(or roots) of the equation.

Hope it helps!

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