A quadratic equation in the form of
a x² + b x + c=0, has
Discriminant = D= b²- 4 ac
x = [tex]\frac{-b \pm \sqrt{D}}{2a}[/tex]
Now coming to roots of a quadratic equation
1. D≥0, both the roots are real i.e both of them may be rational or both of them may be rational.
2. D=0, both the roots are real and equal.
3. D< 0, both the roots are imaginary.
So, out of following options given,
Option A, is not true,
zero, because the discriminant is negative.(It is a true statement if you are talking about real roots but if we consider imaginary roots also then this statement becomes false.) .In the question it is given that that roots are in radical form that's why this option is incorrect.
. Option B is not true because if Discriminant is not a perfect square then also the quadratic function has two real either rational or irrational roots.
Third option is completely false , it is incorrect statement.[one, because the negative and the minus cancel each other]
Fourth option is true because , in the answer it has been written that the roots are in simplest radical form , The value of D should always be greater than zero,then we look at ± symbol .Then there are two possible real roots.
The value of D should always be greater than zero, then we look at the ± symbol. Then there are two possible real roots.
The equation which has the highest power or degree is 2 called the quadratic equation.
The general form of the quadratic equation is given by;
[tex]\rm ax^2+bx+c=0[/tex]
Where; a, b, and c are the constants and a can not be zero.
The discrimination for the quadratic equation is given by;
[tex]\rm Discriminate = b^2-4ac[/tex]
Hence, the value of D should always be greater than zero, then we look at the ± symbol. Then there are two possible real roots.
To know more about the Quadratic equation click the link given below.
https://brainly.com/question/3751209
