Question

If the coefficient of ${x^2}$ and the constant term of the quadratic equation have the opposite sign then the quadratic equation has roots which are:
A. real and distinct
B. real and equal
C. imaginary
D. none of the above

Answer 1

Hint:
For the given quadratic equation of the form $a{x^2} + bx + c = 0$the roots are real if $D \geqslant 0$ and if the roots are imaginary then $D < 0$ and if $D = 0$ then the roots are real and equal and here $D$ represents the discriminant.

Complete step by step solution:
Now here we are given that the coefficient of ${x^2}$ and the constant term of the quadratic equation have the opposite signs.
So let us assume that the quadratic equation be $a{x^2} + bx + c = 0$ here $a$ is the coefficient of ${x^2}$ and $c$ is the constant term.
Here it is given that $a,c$ have the opposite signs that means that if a is positive then c will be negative and if a is negative then c is positive
Now for real root as we know that $D$ must be equal to zero
So $D = {b^2} - 4ac$
And if $
  a > 0,c < 0 \\
  a < 0,c > 0 \\
 $
So product $ac$ will always have the negative sign and we know that ${b^2}$ will always be positive because squares can never be negative.
Therefore $D = {b^2} - 4ac$
Here ${b^2}$ is positive and $4ac$ is negative and therefore D will be positive
So the roots cannot be imaginary they are real and distinct.
Let $\alpha ,\beta $ are the roots of the quadratic equation $a{x^2} + bx + c = 0$
Then the products of the roots$ = \dfrac{c}{a}$
And we know that when $
  a > 0,c < 0 \\
  a < 0,c > 0 \\
 $
So $\dfrac{c}{a} < 0$ as a and c are off opposite signs
So we get that $\alpha \beta < 0$

So as both the roots are of opposite signs they are real and distinct.

Note:
If we are given that the quadratic equation $f(x)$ is always increasing where $f(x) = $$a{x^2} + bx + c = 0$ and if $f(x)$ is increasing then its derivative must also be positive for every value of $x$ that is $f(x) > 0$ for $2ax + b > 0$
So we can write that $x > \dfrac{{ - b}}{{2a}}$
This tells us that if $f(x)$ is increasing then $x > \dfrac{{ - b}}{{2a}}$