Sign of Quadratic Expression - Practice Questions & MCQ

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,

• a + b + c = a.1² + b.1 + c = f(1)
• 9a + 3b + c = a.3² + b.3 + c = f(3)
• 4a - 2b + c = a.(-2)² + b.(-2) + c = f(-2)
• c = a.0² + b.0 + c = f(0)
• 8a - 4b + 2c = 2(4a - 2b + c) = 2.f(-2)