For this case we have that by definition, the discriminant of a quadratic equation, [tex]ax ^ 2 + bx + c = 0[/tex], is given by:
[tex]d = b ^ 2-4 (a) (c)[/tex]
We have to:
[tex]d> 0[/tex]: Two different real roots
[tex]d = 0[/tex]: Two equal real roots
[tex]d <0[/tex]: Two different complex roots
Thus, the answer is option C.
ANswer:
Option C
There will be two real roots or solutions when [tex]b^{2} -4ac[/tex] is positive.
Let us say the quadratic equation is
[tex]ax^{2} +bx+c=0[/tex]......(1)
A quadratic equation is a single-variable polynomial equation of second degree.
The solution of the above equation (1) is given by:
[tex]x= \frac{-b+-\sqrt{b^{2} -4ac} }{2a }[/tex].....(2)
Where [tex]b^{2} -4ac[/tex] is called the discriminant of a quadratic equation.
Suppose [tex]b^{2} -4ac[/tex] is negative [tex]\sqrt{b^{2} -4ac}[/tex] will be imaginary,
So in this case, roots also will be imaginary.
If, [tex]b^{2} -4ac[/tex] is positive [tex]\sqrt{b^{2} -4ac}[/tex] will be a real number,
So in this case, roots will be real and distinct because in (2) we will get two roots because of the ± sign.
Therefore, there will be two real roots or solutions when [tex]b^{2} -4ac[/tex] is positive.
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