Answer :
Answer:
Roots are real, irrational and unequal
Step-by-step explanation:
Consider the Quadratic equation of the form:
[tex]ax^{2} +bx+c=0[/tex]
Roots of above equation are:
[tex]x=\frac{(-b+\sqrt{b^{2}-4ac }) }{2a}[/tex]
[tex]x=\frac{(-b-\sqrt{b^{2}-4ac }) }{2a}[/tex]
So,Discriminant=[tex]b^{2}-4ac[/tex]
Conditions:
Discriminant =24>0
Discriminant =24 is not a perfect square
So looking at the above conditions we can say:
Roots are real, irrational and unequal
The roots of a quadratic equation whose discriminant is 24 are real, irrational, and distinct.
Given that:
Discriminant of a quadratic equation: 24
Explanation and calculations:
The roots of a quadratic equation [tex]ax^2 + bx + c= 0[/tex] is given as:
[tex]x = \dfrac{-b \pm \sqrt{D}}{2a}\\\\D = b^2 - 4ac[/tex]
When discriminant is positive, that means roots are real, else if they're negative, then that means the roots are imaginary. If the discriminant is 0, then roots are equal.
The given quadratic equation's discriminant is 24, thus the roots are real.
[tex]\sqrt{24} = 2\sqrt{6} \equiv \rm irrational[/tex]
Thus the roots are also irrational as -b/2a is rational; and an irrational number added or subtracted with rational numbers gives irrational result.
The roots are distinct since discriminant is not 0.
Thus, the roots of a quadratic equation whose discriminant is 24 are real, irrational and distinct.
Learn more about roots(also called zeros) of quadratic equations here:
https://brainly.com/question/15884086