Question

The cubic equation whose roots are the squares of the roots of $x^3-2x^2+10x-8=0$ is:

• A. $x^3+16x^2+68x-64=0$
• B. $x^3+8x^2+68x-64=0$
• C. $x^3+16x^2-68x-64=0$
• D. $x^3-16x^2+68x-64=0$

Answer 1

Answer: (A)

$x^3+16x^2+68x-64=0$

Let $x^3-2x^2+10x-8=0$ has roots $a,b,c.$

$\therefore a+b+c=2,ab+bc+ca=10,abc=8$

The new roots would be $a^2,b^2,c^2$

$\Rightarrow a^2+b^2+c^2=(a+b+c{)}^2-2(ab+bc+ca)=2^2-2(10)=-16$

$a^2{b}^2+b^2{c}^2+a^2{c}^2$

$=(ab+bc+ac{)}^2-2(a{b}^{2}c+b{c}^{2}a+a^{2}bc)$

$=(ab+bc+ac{)}^2-2abc(a+b+c)$

$={10}^2-2\left(8\right)\left(2\right)=100-32=68$

$a^2{b}^2{c}^2=(abc{)}^2=8^2=64$

The new equation can thus be written as $x^3-(-16)x^2+68x-64=0$

i.e. $x^3+16x^2+68x-64=0$