Question

# If the zeros of the quadratic polynomial $\text{a}{{\text{x}}^{\text{2}}}\,\text{+}\,\text{bx}\,\text{+}\,\text{c,}\,\text{c}\,\ne \,\text{0}$ all equal, thenA) c and a have opposite signsB) c and b have opposite signsC) c and a have the same signsD) c and b have the same signs

Hint: As you can see in the question $\text{a}{{\text{x}}^{\text{2}}}\,\text{+}\,\text{bx}\,\text{+}\,\text{c,}\,\text{c}\,\ne \,\text{0}$ then it is clear that discriminant of the polynomial is zero using the equation for discriminant $\text{(i}\text{.e}\,{{\text{b}}^{\text{2}}}\,\text{-}\,\text{4ac)}$ we can further find which variables have same sign.

Complete step by step solution:
Now as given in the question that the zeros of polynomial $\text{a}{{\text{x}}^{\text{2}}}\,\text{+}\,\text{bx}\,\text{+}\,\text{c,}\,\text{c}\,\ne \,\text{0}$ are equal.
So the value of discriminant ‘D’ has to be zero.
$\therefore$ the equation for discriminant is,
${{\text{b}}^{\text{2}}}\,-\,\text{4ac}\,\text{=}\,\text{0}$
Now, shifting the term $4\text{ac}$
${{\text{b}}^{\text{2}}}\,=\,\text{4ac}$
We have to analyze one thing at this point that is the term $'{{\text{b}}^{\text{2}}}'$ on LHS can never be negative because the square of any number is always positive.
This means that the term on RHS i.e $4\text{ac}$ is also always positive.
Now, the term $'4\text{ac }\!\!'\!\!\text{ }$ to be always positive, it is necessary that both ‘a’ & ‘c’ should have the same sign (i.e either they both should be positive or both should be negative).

$\therefore$ Option (C) both c and a have the same sign is correct.

Note: A discriminant can be positive, zero or negative, and this determines how many solutions three are to a given quadratic equation. A positive discriminant indicates that quadratic has two distinct real roots. A discriminant zero indicates that the quadratic has a repeated real number solution.