# Thread: further maths roots of equations

1. ## further maths roots of equations

Hi can any body solve this question?

The equation 2x²-4x-3=0 has roots a and b. Find the quadratic equation with integer coefficients and with roots a³ , b³

2. Hello,

Using Viete, you know the values of a+b and ab.
Since a^3+b^3=(a+b)^3-3ab(a+b), you know the values of p=a^3+b^3 and q=a^3b^3.
The quadratic equation with roots a^3, b^3 is x^2-px+q=0.
Arrange it so that it has integer coefficients.

Bye.

3. Hello, djmccabie!

The equation $2x^2-4x-3\:=\:0$ has roots $a$ and $b.$
Find the quadratic equation with integer coefficients and with roots $a^3,\:b^3$
The given equation is: . $x^2 - 2x - \frac{3}{2} \:=\:0$ with roots $a$ and $b.$

Hence: . $\begin{array}{cccc}a + b &=& 2 & {\color{blue}[1]} \\ \\[-4mm] ab &=&\text{-}\frac{3}{2} & {\color{blue}[2]} \end{array}$

Cube [1]: . $(a + b)^3 \:=\: 2^3 \quad\Rightarrow\quad a^3 + 3a^2b+3ab^2 + b^3 \:=\:8$

$\text{We have: }\;(a^3 + b^3) + 3\underbrace{(ab)}_{-\frac{3}{2}}\underbrace{(a + b)}_{2} \;=\;8$

. . Hence: . $a^3+b^3 - 9 \:=\:8\quad\Rightarrow\quad \boxed{a^3+b^3 \:=\:17}$

Cube [2]: . $(ab)^3 \:=\:\left(-\frac{3}{2}\right)^3 \quad\Rightarrow\quad\boxed{ a^3b^3 \:=\:-\frac{27}{8}}$

Therefore, the equation is: . $x^2 - 17x - \frac{27}{8}\:=\:0 \quad\Rightarrow\quad \boxed{8x^2 - 136x - 27 \:=\:0}$

4. Thanks alot soroban! really helped i had the right idea just got a few signs wrong