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
Hello, djmccabie!
The given equation is: .$\displaystyle x^2 - 2x - \frac{3}{2} \:=\:0$ with roots $\displaystyle a$ and $\displaystyle b.$The equation $\displaystyle 2x^2-4x-3\:=\:0$ has roots $\displaystyle a$ and $\displaystyle b.$
Find the quadratic equation with integer coefficients and with roots $\displaystyle a^3,\:b^3$
Hence: .$\displaystyle \begin{array}{cccc}a + b &=& 2 & {\color{blue}[1]} \\ \\[-4mm] ab &=&\text{-}\frac{3}{2} & {\color{blue}[2]} \end{array}$
Cube [1]: .$\displaystyle (a + b)^3 \:=\: 2^3 \quad\Rightarrow\quad a^3 + 3a^2b+3ab^2 + b^3 \:=\:8$
$\displaystyle \text{We have: }\;(a^3 + b^3) + 3\underbrace{(ab)}_{-\frac{3}{2}}\underbrace{(a + b)}_{2} \;=\;8$
. . Hence: .$\displaystyle a^3+b^3 - 9 \:=\:8\quad\Rightarrow\quad \boxed{a^3+b^3 \:=\:17}$
Cube [2]: .$\displaystyle (ab)^3 \:=\:\left(-\frac{3}{2}\right)^3 \quad\Rightarrow\quad\boxed{ a^3b^3 \:=\:-\frac{27}{8}}$
Therefore, the equation is: .$\displaystyle x^2 - 17x - \frac{27}{8}\:=\:0 \quad\Rightarrow\quad \boxed{8x^2 - 136x - 27 \:=\:0}$