Question

The quadratic equation with rational, coefficient the sum of the squares of whose roots is $40$ and the sum of the cubes of whose root is $208$ is

• A. $x^2+4x+12=0$
• B. $x^2–4x–12=0$
• C. $x^2–4x+12=0$
• D. $x^2+4x–12=0$

Answer 1

The correct option is B

$x^2–4x–12=0$

Find the quadratic equation using the sum of the squares of the roots and sum of the cubes of the roots.

Let $ɑ,\beta$ be the roots.

Given that,

${ɑ}^2+{\beta }^2=40$,

${ɑ}^3+{\beta }^3=208$

Let us suppose that

$ɑ+\beta =m$, $\alpha \beta =n$

we know that

$(a^2+b^2-ab)(a+b)=a^3+b^3$

$({(a+b)}^2-(a^2+b^2\left)\right)/2=ab$

so,

$$\begin{gathered} (ɑ+\beta )({ɑ}^2+{\beta }^2–ɑ\beta )=208 \\ m(40–n)=208 \\ m\left[\right(40)–(m^2–40)/2]=208 \\ {m}^3–120m+416=0 \\ (m–4)(m^2+4m–104)=0 \\ m–4=0 \\ m=4 \\ n=[m^2–40]/2=–12 \end{gathered}$$

We know that

if $a{x}^2+bx+c$is the quadratic equation, $p$ and $q$ be the roots of the equation.

then $p+q=-b/a$, $p\times q=c/a$

so,

$x^2–4x–12=0$

Hence the quadratic equation with rational, coefficient the sum of the squares of whose roots is $40$ and the sum of the cubes of whose root is $208$ is $x^2–4x–12=0$