CAT 2021 Slot 2 · Quant · Question 19

Because the coefficients are rational, irrational roots come in conjugate pairs, so the other root is 2-\sqrt{3}.

For a{x}^{2}-bx+c=0 the sum of the roots is \frac{b}{a} and the product is \frac{c}{a}.

Sum: \left( 2+\sqrt{3}\right)+\left( 2-\sqrt{3}\right)=4,

so b=4a.

Product: \left( 2+\sqrt{3}\right)\left( 2-\sqrt{3}\right)=4-3=1,

so c=a.

Now apply b={c}^{3}: 4a={a}^{3},

so a\left({a}^{2}-4 \right)=0.

Since a\ne 0, a=\pm 2 and \left| a \right|=2.