Quadratic Inequalities

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Equation of form ax2 + bx + c = 0 where a, b and c are real numbers is called quadratic equations. If x=α is a root of the equation, then f(α)=0

The root of the quadratic equation of form ax2 + bx + c = 0 is 

x = \[\frac{-b  ±  \sqrt{b^{2}-4ac}}{2a}\]

Here D is called the Discriminant of the quadratic expression and is equal to 

D = b2 - 4ac 

The nature of roots depends upon the value of D 

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  1. If D < 0, then roots are non-real and complex 

  2. If D > 0, then roots are real and distinct 

  3. If D = 0, then roots are real and equal 

Steps to Solving Quadratic Inequalities

  1. Factorise the quadratic equation by putting ax2 + bx + c = 0 

  2. Determine the critical values i.e. values at which the expression becomes zero

  3. Plot these critical values and assign sign using the wavy curve method

  4. Plot a rough sketch or graph 


Solving Quadratic Inequalities

a. Check the Discriminant of Quadratic Expression, f(x) = ax2 + bx + c; then 

    1. If D > 0, then the expression can be written as f(x) = a (x – α)(x – β). where α and β are given by

    2. If D = 0, then the expression can be written as  f(x)=c(x−α)2

    3. ​If D > 0 & if A > 0, then f(x) > 0 for all ∀ X ϵ R and the expression will be cross multiplied and the sign of the inequality will not change.

  • If A < 0, then f(x) < 0  for all X ϵ R and the expression will be cross multiplied and the sign of the inequality will change.

  • If the expression (say ‘f’) is cancelled from the same side of the inequality, then cancel it and write f ≠ 0 e.g.

Graph of Quadratic Expression

Vertex of parabola y = ax2 + bx + c

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Vertex of the parabola is (-b/2a,-D/4a)

If a > 0 then parabola opens upward 

If a < 0 graph opens downward


Intersection with Axis

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  1. If D > 0, parabola cuts the x-axis in two real and distinct points and opens upward if a > 0 while opens downward if a < 0

  2. If D = 0 graph touches the x-axis and opens upward if a > 0 while opens downward if a < 0

  3. If D < 0 graph does not touch the x-axis i.e. has no real roots and opens upward if a > 0 while opens downward if a < 0


Maximum and Minimum values of y = ax2 + bx + c 

For a > 0, the parabola y = f(x) opens upward and has minimum value at x = -b/2a 

f(-b/2a) = -D/4a

Range=(-D/4a,∞)

For a < 0, the parabola y = f(x) opens upward and has maximum value at x = -b/2a 

f(-b/2a) = -D/4a

Range=(∞,-D/4a)


Sign of Quadratic Equation

1. If b2 - 4ac = 0 (double root case), then we have

ax+ bx + c = a(x+b/2a)2

In this case, the function f(x) = ax2 + bx + c has the sign of the coefficient a.

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2. If b2 - 4ac > 0 (two distinct real roots case). In this case, we have

                           ax2 + bx + c = a(x - x1)(x - x2)

Where x1 and x2 are the two roots with x1 < x2 Since (x - x1)(x - x2) is always positive when x < x1and x > x2, and always negative when x1 < x< x2 , we get

  • ax2 + bx + c has same sign as the coefficient a when x<x1and x>x2;

  • ax2 + bx+ c has the opposite sign as the coefficient a when x1 < x< x2 .

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3. If b2 < 4ac (complex roots case),ax2 + bx + c then has a constant sign the same as the coefficient a.

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Wavy Curve Method

  1. Factorize the  given polynomials

  2. Now make the coefficient of all the variable to factors positive

  3. Multiply/divide the equation both sides of the inequality by -1 remove the minus sign and by doing the inequality will reverse

  4. Find the roots and asymptotes of the given inequality by equating each of the factors to 0

  5. Now finally Plot the points on the number line and start with the largest factor. Here the curve from the positive region of the number line should intersect that point. Now let's look at the power of the factors and If it is odd, then we have to change the path of the curve from their respective roots while if it's even, continue in the same region.

FAQ (Frequently Asked Questions)

Q1. Solve the given Quadratic Inequality x2 – 4x > –3

Ans. Step 1: Bring all terms on one side making other side of the inequality zero

x2 – 4x > –3

x2 – 4x + 3 > 0

Step 2: Factorise the given quadratic expression

x2 – 4x + 3 > 0

(x – 3)(x – 1) > 0

Step 3: Now find the range of values of x which satisfies the given inequality.

(x – 3)(x – 1) > 0 (y is positive): We choose the interval for which the curve is above the x-axis.

x < 1 or x > 3

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