Question

Let $a,b,c,d$ be four integers such that $ad$ is odd and $bc$ is even, $a{x}^3+b{x}^2+cx+d=0$ has

• A. at least one irrational root
• B. all these rational roots
• C. all these integral roots
• D. none of these

Answer 1

Answer: (A)

at least one irrational root

Let us assume all the roots are rational !

ad is odd, therefore, a is odd as well as d is odd

bc is even, therefore, either of them or both are even

We can write $f\left(x\right)=a{x}^3+b{x}^2+cx+d=(a_{1}x+a_2)(b_{1}x+b_2)(c_{1}x+c_2)$

Where $a_1,b_1,c_1,a_2,b_2$ and $c_2$ all are integers!

$f\left(x\right)=a_1{b}_1{c}_1{x}^3+(a_1{b}_1{c}_2+a_1{b}_2{c}_1+a_2{b}_1{c}_1)x^2+(a_1{b}_2{c}_2+a_2{b}_1{c}_2+a_2{b}_2{c}_1)x+a_2{b}_2{c}_2$

Now we can say,

$a=a_1{b}_1{c}_1$

$b=a_1{b}_1{c}_2+a_1{b}_2{c}_1+a_2{b}_1{c}_1$

$c=a_1{b}_2{c}_2+a_2{b}_{1}c2+a_2{b}_2{c}_1$

$d=a_2{b}_2{c}_2$

As already mentioned above that a and d both are odd,

Therefore, $a_1,b_1,c_1,a_2,b_2$ and $c_2$ all are ODD

This information tells us that b and c both are also ODD

As I said either of b or c is EVEN

This is our CONTRADICTION !!!

Therefore our assumption was wrong and all roots are NOT rational.

Hence it has at least one irrational root.