Quadratic Equation and Inequations

QUADRATIC EQUATIONS AND INEQUATIONS

Quadratic equations:

The standard form of a quadratic equation is ax²+bx+c = 0 where a, b and c are known values and a ≠0. Also, x is a real variable.

If the value of a is 0, then the equation will become a linear equation.

(i) The Discriminant of a quadratic equation is ax² + bx + c = 0 (a ≠ 0) is 

∆ = b² - 4ac

(ii) If α and β be the roots of the equation ax² + bx + c = 0 (a ≠ 0) then

α + β =  -b/a = -coefficient of x/coefficient of x² 

and αβ = c/a = constant term/coefficient of x²

(iii) The formula for the formation of the quadratic equation whose roots are given: x² - (sum of the roots)x + product of the roots = 0.

 

Example 1: Solve for x: x²-3x-10 = 0

Solution:

Let us express -3x as a sum of -5x and +2x.

→ x²-5x+2x-10 = 0

→ x(x-5)+2(x-5) = 0

→ (x-5)(x+2) = 0

→ x-5 = 0        or         x+2 = 0

→ x = 5           or         x = -2

 

Example 2: Solve for x: 11x²+18x+7 = 0

Solution:

In this case, the sum of the numbers we choose should equal to 18 and the product of the numbers should equal 11*7 = 77.

This can be done by expressing 18x as the sum of 11x and 7x.

→ 11x²+11x+7x+7 = 0

→ 11x(x+1) +7(x+1) = 0

→ (x+1)(11x+7) = 0

→ x+1 = 0       or         11x+7 = 0

→ x = -1          or         x = -7/11.

 

Example 3 :Solve for x: x² = 24 – 10x

Solution:

Rewriting the equation into the standard quadratic form,

X² +10x-24 = 0

What are the two numbers which when added give +10 and when multiplied give -24? 12 and -2.

So this can be solved by the factoring method. But let’s solve it using the new method, applying the quadratic formula.

Here, a = 1, b = 10 and c = -24.

x = [-10 ± √(100 – 4*1*-24)] / 2*1

x = [-10 ± √(100-(-96))] / 2

x = [-10 ± √196] / 2

x = [-10 ± 14] / 2

x = 2 or x= -12 are the roots.

 

Discriminant

For an equation ax²+bx+c = 0, b²-4ac is called the discriminant and helps in determining the nature of the roots of a quadratic equation.

If b²-4ac > 0, the roots are real and distinct.

If b²-4ac = 0, the roots are real and equal.

If b²-4ac < 0, the roots are not real (they are complex).

 

Example 4: Find the nature of roots for the equation x²+x+12 = 0.

Solution:

b²-4ac = -47 for this equation. So it has complex roots. Let us verify this.

→ [ -1±√(1-48)] / 2(1)

→ [-1±√-47] / 2

√-47 is usually written as i √47 indicating it’s an imaginary number.

Hence verified.

 Inequations:

 As in so many cases, Inequations are known as Inequalities.

In these questions, you will be provided with a statement consisting of a group of elements. These elements will be having a certain coded relationship among them which is denoted by different inequality symbols like >, <, =, etc.

S.No          Symbol                  Meaning

1              A > B              A is Greater than B.

2.             A < B              A is Smaller than B

3.             A = B               A is Equals to B.

4.             A ≠ B             A is either greater than or smaller than B

5.             A ≥  B            A is Greater than or Equals to B

6.             A ≤ B              A is Smaller than or Equals to B

Example 1: Solve the inequality 2x+1/x+3 < 1

Solution: We have 2x+1/x+3< 1

Here the right-hand side is not equal to zero. So transpose 1 to the left-hand side and solve. We get 2x+1/x+3 < 1

⇒ (2x+1/x+3)-1 < 0

⇒ (2x+1-x-3/x+3) <0

⇒ (x-2/x+3) < 0

The critical points are x = 2, -3. Plot these points on the number line, we get

Since the given inequality is negative, so the solution is -3 < x < 2

 

Example 2: Solve |9x – 4| - 5 < 11.

Solution: We have |9x – 4| - 5 < 11

⇒ |9x – 4| < 11 + 5

⇒ |9x – 4| < 16

⇒ - 16 < 9x – 4 < 16

⇒ - 16 + 4 < 9x < 16 + 4

⇒ -12 < 9x < 20 ⇒ (-12/9) < x < (20/9)

⇒ (-4/3) < x < (20/9) , which is the required solution.

Example 3: 2x²-5x+2<=0

=> 2x²-4x-x+2<=0

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

=>(x-(1/2))(x-2)<=0

x belongs to [(1/2),2]