# How do I form this equation?

If $A$ and $B$ are the root of the equation $3x^2-4x-9=0$, what is the equation whose roots are $(A+3)/(A-3)$ and $(B+3)/(B-3)$

-
Note that $\frac{A+3}{A}-3 = \frac{3}{A}-2$, and likewise for $B$. Also note that "$3x^2-4x-9$" is not an equation, but a polynomial. You can make it into an equation by appending "$=0$". – Henning Makholm Jul 11 '12 at 1:00
Did you mean $\dfrac{A+3}A - 3$ or $\dfrac{A+3}{A - 3}$? – user17762 Jul 11 '12 at 1:02
the latter one, sorry – HackToHell Jul 11 '12 at 1:05

Recall that if $\alpha, \beta$ are roots of $ax^2 + bx + c$, then we have $\alpha + \beta = -\dfrac{b}a$ and $\alpha \beta = \dfrac{c}a$.

Let $\alpha = \dfrac{A+3}{A-3}$ and $\beta = \dfrac{B+3}{B-3}$.

Then we get that $$\alpha + \beta = \dfrac{A+3}{A-3} + \dfrac{B+3}{B-3} = \dfrac{AB + 3B - 3A - 9 + AB + 3A - 3B - 9}{AB - 3(A+B) + 9} = 2\dfrac{AB - 9}{AB - 3(A+B) + 9}$$

Similarly, $$\alpha \beta = \dfrac{AB + 3(A+B) + 9}{AB - 3(A+B) + 9}$$ But since $A$ and $B$ are roots of $3x^2 -4x - 9$, we have that $A+B = \dfrac43$ and $AB = -3$.

Can you finish it off from here?

Move the mouse over the gray area for the complete solution.

$$\alpha + \beta = 2 \left(\dfrac{AB - 9}{AB - 3(A+B) + 9} \right) = 2 \times \dfrac{-3-9}{-3 - 3(4/3) + 9} = 2 \times \dfrac{-12}{2} = -12$$ $$\alpha \beta = \dfrac{-3 + 3 (4/3) + 9}{-3 - 3 (4/3) + 9} = \dfrac{10}2 = 5$$ Hence, the desired quadratic polynomial is $$y^2 + 12y + 5.$$

-
Yes thanks for helping – HackToHell Jul 11 '12 at 1:07

Remember that the quadratic polynomial with roots $r$ and $s$ is $(x-r)(x-s) = x^2-(r+s)x + rs$.

You know that $(x-A)(x-B) = x^2 - \frac{4}{3}x - 3$, so $AB =-3$ and $A+B = \frac{4}{3}$. From these, you should be able to find a quadratic with the roots you want by expressing the fum and the product of $$\frac{A+3}{A-3} \qquad\text{and}\qquad \frac{B+3}{B-3}$$ in terms of $A+B$ and $AB$.

-

More generally, suppose $p(x)$ is a polynomial with roots $r_j$. Note that $y = \dfrac{ax+b}{cx+d}$ (where $ad - bc \ne 0$) when $x = \dfrac{dy - b}{-cy + a}$. So $\dfrac{a r_j+b}{c r_j+d}$ (assuming the denominators are nonzero) are roots of $p\left( \dfrac{dy-b}{-cy+a} \right)$. In this case with $p(x) = 3 x^2 - 4 x - 9$ and $a=1$, $b=3$, $c=1$, $d=-3$, $$p\left( \dfrac{-3 y - 3}{-y + 1}\right) = 3{\frac { \left( -3\,y-3 \right) ^{2}}{ \left( -y+1 \right) ^{2}}}-4 \,{\frac {-3\,y-3}{-y+1}}-9 = 6\,{\frac {{y}^{2}+12\,y+5}{ \left( y-1 \right) ^{2}}}$$ and the roots of this are the roots of $y^2+12y+5$. This method works with polynomials of arbitrary degree, not just quadratics.

-
Thanks for the general equation ... – HackToHell Jul 11 '12 at 14:10

A+B = 4/3

AB = -3

(A+3)/(A-3) + (B+3)/(B-3) = 2(AB - 9)/[AB - 3(A+B) + 9]

= 2(-3 - 9)/[-3 - 3(4/3) + 9] = -12

[(A+3)/(A-3)][(B+3)/(B-3)] = [(A+3)(B+3)]/[(A-3)(B-3)] = [AB + 3(A+B) + 9]/[AB - 3(A+B) + 9]

= [-3 + 3(4/3) + 9]/[-3 -3(4/3) + 9]

= [-3 + 4 + 9]/[-3 - 4 + 9]

= 10/2

= 5

The new equation is framed below.

x^2 - (-12)x + (5) = 0

x^2 + 12x + 5 = 0

-
An equation involving just one variable x in second degree is of the form ax^2 + bx + c = 0 – Rajesh K Singh Jul 11 '12 at 5:06
A second degree equation involving just one variable x with roots a and b is framed as follows. x^2 - (a + b)x + (a)(b) = 0 – Rajesh K Singh Jul 11 '12 at 5:17