Chapter 7.4, Problem 4ST

To determine

Without solving, find the nature of the roots.

Answer to Problem 4ST

two rational and different roots.

Explanation of Solution

Given:

The equation is :

  $3y^2-9y=0$

Formula Used:

For the quadratic equation $a{x}^2+bx+c=0$ , the determinant D describes the nature of the roots :

  $D=b^2-4ac$

If D >0 : If it is perfect square , the two roots are different and rational ; if it is not a perfect square , the two roots are different and irrational .

If D = 0 , then the roots are real and equal .

If D < 0 , then the roots are imaginary and conjugate.

Calculation:

The determinant of $3y^2-9y=0$ :

Here a = 3 , b = -9 , c = 0

$D=b^2-4ac={\left(-9\right)}^2-4\cdot 3\cdot (0)=81+0=81>0$

81 is a perfect square of 9 .

So, the roots are two different and rational.