Question

A quadratic equation with rational coefficients can have

• A. one root real, other imaginary
• B. one root rational, other irrational
• C. both negative and irrational roots
• D. have one non-real and other real root

Answer 1

Answer: (C)

both negative and irrational roots

$a{x}^2+bx+c=0$

where $a,b,c$ are rationals

The roots of the above equation are given by the quadratic formula

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Case I

$b^2-4ac<0$

Then $x=\frac{-b\pm i\sqrt{|b^2-4ac|}}{2a}$

Thus, both roots are imaginary.

Case II

$b^2-4ac=0$

Then the roots are equal and either positive or negative.

Case III

$b^2-4ac>0$

$\sqrt{b^2-4ac}>b$

Then the roots are real and unequal.

Case IV : $b^2-4ac>0$ and perfect square

Then the roots are real, rational and unequal.

If the coefficients are rational, then it is not possible to have one imaginary and one real root.

Case V : $b^2-4ac>0$ and not a perfect square

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

and $b>\sqrt{b^2-4ac}$

$⟹$ Roots are negative, irrationals and unequal.

Hence, option C is correct.

a≠ 0 and discriminant is positive (i.e., b