Question

is this proof correct for the below mentioned question

prove that for a quadratic equation f(x)=ax^2+bx+c, the discrminant will be negative if roots of f(x) are imaginary


PLEASE DO CHECK​

Answer 1

Hi !
I suppose the first statement you've written is absolutely incorrect.
This is because, it Is not necessary that the sum of the two complex conjugate roots will be negative only because it has not been mentioned that b is negative.

However, when I had a look at the proof, it seems pretty unsatisfying but don't worry that was a great try.
Now let me give you a hint on how you prove that if Discriminant is negative then roots are imaginary.

Look Discriminant < 0
Now, By Sridharacharya formula to find roots of a quadratic equation of form ax^2 + bx + c = 0
is
x = -b +- √Discriminant / 2a

So, you see if Discriminant is negative, then the value of x cannot be real because in the real world square roots of negative numbers are an offence until you learn what are complex numbers.

Now there's another way
Geometrically, quadratic equations represent parabola

If a quadratic equation has positive discriminant then it intersects the x - axis at two points which are the roots of the equation.
However, if an equation has negative discriminant, then it would never intersect the x axis meaning that the equation doesn't have any real roots.
But from the theory of polynomials we know that any equation has the number of roots equal to it's degree.
So, we know though there aren't two real roots but two complex conjugate roots for sure !
Hope this helps you !

Answer 2

Answer:

suppose the first statement you've written is absolutely incorrect.

This is because, it Is not necessary that the sum of the two complex conjugate roots will be negative only because it has not been mentioned that b is negative.

However, when I had a look at the proof, it seems pretty unsatisfying but don't worry that was a great try.

Now let me give you a hint on how you prove that if Discriminant is negative then roots are imaginary.

Look Discriminant < 0

Now, By Sridharacharya formula to find roots of a quadratic equation of form ax^2 + bx + c = 0

is

x = -b +- √Discriminant / 2a

So, you see if Discriminant is negative, then the value of x cannot be real because in the real world square roots of negative numbers are an offence until you learn what are complex numbers.

Now there's another way

Geometrically, quadratic equations represent parabola

If a quadratic equation has positive discriminant then it intersects the x - axis at two points which are the roots of the equation.

However, if an equation has negative discriminant, then it would never intersect the x axis meaning that the equation doesn't have any real roots.

But from the theory of polynomials we know that any equation has the number of roots equal to it's degree.

So, we know though there aren't two real roots but two complex conjugate roots for sure !

Hope this helps you !