# Quadratic Inequations

Inequations of the type a x² +b x +c > 0, a x² + b x +c 0, a x² +b x +c < 0, a x² +b x +c 0 (a0) are called quadratic inequalities.

### Solving Quadratic Inequalities

First make the coefficient of x² unity by multiplying throughout by 1/a, and reversing the inequality sign if a is negative. Now, three cases arise:

**Case I.** If the quadratic has real distinct roots, say
, and
< , then

- (x -)(x -) > 0 => x < or x >
- (x - (x -) 0 => x or x
- (x -)(x -) < 0 => < x <
- (x -)(x -) 0 => x

**Case II.** If the quadratic has two real, identical roots say
, , then we know that x²
0 for real x. Therefore

- (x -)² 0 => x R (all real numbers)
- (x -)² > 0 => x R except x =
- (x -)² 0 => x =
- (x -)² < 0 => there is no solution.

**Case III.** If the quadratic has complex roots, then either there is
no solution or the entire set of real numbers is the solution.

### Sign of Quadratic expressions

- a x² +b x +c has same sign as a except when a x² +b x +c = 0 has two real and distinct roots, say , ( < ) and < x < .
- a x² +b x +c is +ve for all real x, when a > 0 and < 0.
- a x² +b x +c is -ve for all real x, when a < 0 and < 0.

### Range (Maximum / Minimum Value) of Quadratic Expressions

Hence the maxima / minima occurs at x = -b/2a and is equal to - /4a. If a > 0, we have a minima, otherwise we have a maxima.

## Illustrative Examples

### Example

Solve algebraically:

- 2 x² +x -1 < 0
- x² +2x +3 0
- x² +x +1 < 0

### Solution

- 2 x² +x -1 < 0 => (x +1)(2 x -1) < 0

=> (x +1)(x -1/2) < 0 (Dividing by 2)

=> -1 < x < 1/2, which is the required solution.

We can also write it as x (-1, 1/2) - For x² +2 x +3 0, the discriminant

= 2² -4.1.3 = -8 < 0, so roots are complex.

Now x² +2 x +3 = (x +1)² +2 2 for all real x.

Hence x² +2 x +3 0 is true for all real x. - For x² +x +1 < 0, discriminant

= 1² -4.1.1 = -3 < 0, so roots are complex.

Now for all real x.

So x² +x +1 < 0 has no solution.

## Exercise

- Discuss the sign of x² -3 x +2.
- Solve the following inequalities:

(i) -x² +3 x -2 > 0 (ii) 4 x² -9 0

(iii) 2 x² +x -15 0. - Solve the following inequalities:

(i) 2 x² < x (ii) 3 -2 x²> 5 x

(iii) x² -6 x +19 0 (iv) x² -6 x +11 < 1. - Find all real values of x which satisfy x² -3 x +2 > 0 and x² -3 x -4 0.
- Find the range of x for which 6 +x < 2 x².
- If x is real, prove that 5 x² -8 x +6 is always positive and find its minimum value.
- If x be real, find the maximum value of a² +2 a x - x².
- Find the values of a so that expression x² -(a +2) x +4 is always positive.
- Construct a quadratic polynomial which is zero when x = 3, 5 and whose minimum value is -1. How many such polynomials are possible?
- Can you construct a quadratic polynomial which is zero when x = 3, 5 and
whose maximum value is

(i) 2 (ii) -2? - Solve: log
_{2}(x² -1) = 3. - Write the most general quadratic expression whose

(i) minimum value is K

(ii) maximum value is K.

## Answers

**1.**-ve when 1 < x < 2, zero at x = 1, 2; +ve otherwise.

**2.**(i) (1, 2) (ii) x > 3/2 or x < -3/2

(iii)

**3.**(i) (ii) -3 < x < 1/2 (iii) all real values of x

(v) no real value of x.

**4.**-1 x < 1 or 2 < x 4.

**5.**x < -3/2 or x > 2.

**6.**14/5

**7.**2a² at x = a.

**8.**-6 < a < 2.

**9.**x² -8 x +15; only one.

**10.**(i) Yes, -2 x² +16 x -30 (ii) No.

**11.**-3 x < -1 or 1 < x 3.

**12.**(i) (a x +b)² +K (ii) K -(a x +b)²