Let y = ax2 + bx + c = 0 be the quadratic equation, such that a is non-zero and a,b,c are real numbers, then
1. If D < 0 then we know quadratic equation has no real roots. So for all real value of x, the graph never intersects or touches x axis, so it is always above or below x-axis. This means that the value of y will always be positive or negative,
If a > 0 and D < 0:
The graph open upwards hence all values of y will be positive as graph can’t start from below x-axis (because if it happens then it will cut x - axis as it is opening upwards, but it has no solution) so it starts from above x-axis and hence y is +ve for all values of x.
In similar way if a < 0 and D < 0 then y is -ve for all values of x- axis.

2. If D = 0, then the quadratic equation y will have one real solution, so y = 0 for one particular value of x and for all rest value of x, y will be +ve or -ve depending upon value of a. If a > 0, then the graph will open upwards so y will be +ve otherwise if a < 0, then y will be -ve.
3. If D > 0, then the quadratic equation y will have two real solution 𝛂 and 𝜷, so if a > 0 then y = 0 on 𝛂 and 𝜷, and between the solution (𝛂 < x < 𝜷),, y will be -ve and left (for x < 𝛂 ) and right (x > 𝜷) part of the solution will give +ve value of y
If a < 0, exactly the opposite will happen, y = 0 on 𝛂 and 𝜷, and between the solution (𝛂 < x < 𝜷), y will be +ve and left (for x < 𝛂 ) and right (x > 𝜷) part of the solution will give -ve value of y.

Note
If f(x) = ax2 + bx + c, then linear expressions can be identified in terms of functions at some constant value
Eg,
| Exam | Chapter |
| JEE MAIN | Complex numbers and quadratic equations |
If $\\\mathrm{f(x) = x^2 + 2(a-1)x + (a+5)}$ , then the values of ‘a’ for which f(x) = 0 have two real and equal roots is
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If $x^2+2(a-1) x+(a+5)>0$ for all $x$ in real numbers then a lies between
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If
then for $x \in (a',b)$ , so the value of b is:
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The values of for which
is a > n, then the value of n is
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Values of for which
are:
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Number of integer values of for which
is:
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The number of integral values of m for which the quadratic expression, is always positive, is :
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If are the length of the sides of a triangle, then r cannot be equal to:
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Let p and q are two positive numbers such that and
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The number of integers which can take so that
is:
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Let , then the value of
for which
is
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If $a x^2+b x+c \geq 0$ for all $c \in R$, where $a=4, b=7$ then c belongs to
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If $a x^2+b x+c \leq \delta$ for $x \epsilon R$; where 'a' is a negative real number and discriminant of the equation is non- positive real number. Then $\delta$ should be
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If $c(4 a-2 b+c)<0$, then the roots of $a x^2+b x+c=0$ are $(a, b, c \in R)$
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The signs of $a \: and\: c$ if the graph of $y=ax^{2}+bx+c$ is as follows

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Find K such that both roots of the quadratic equation $K x^2+x-3=0$ are positive.
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