Question

For the quadratic equation $a{x}^2+bx+c=0$,
Match the given conditions with the correct nature of roots.

• A. Distinct Integers
• B. Non Real - Unequal
• C. Rational - Unequal
• D. Irrational - Unequal

Answer 1

For any quadratic equation $a{x}^2+bx+c=0,\text{discriminant}\left(D\right)=b^2-4ac$

When $a=1,b=10,c=16$,
$D=(10{)}^2-4(1\left)\right(16)=100-64=36=6^2$
Since $a=1\text{and}D$ is a perfect square of an integer, hence the equation will have distinct integers as its roots

When $a=2,b=3,c=-2$,
$D=(3{)}^2-4(2\left)\right(-2)=9+16=25=5^2$
Since $a\ne 1\text{and}D$ is a perfect square of a rational number, hence the equation will have rational and unequal roots.

When $a=1,b=4,c=\frac{9}{4}$
$D=(4{)}^2-4(1)\left(\frac{9}{4}\right)=16-9=7$
Since $D>0$ but it is not a perfect square of a rational number, hence the equation will have irrational and unequal roots.

When $a=1,b=1,c=1$
$D=(1{)}^2-4(1\left)\right(1)=1-4=-3$
Since $D<0$, hence the equation will have non-real and unequal roots.