Quadratic Equation by Completing Square Part - Quadratic Equations, CBSE, Class 10, Mathematics | Extra Documents, Videos & Tests for Class 10 PDF Download

Solution by completing the square

Algorithm :

Step-I : Obtain the quadratic equation. Let the quadratic equation be ax² + bx + c = 0, a ≠ 0.

Step-II : Make the coefficient of x² unity by dividing throughout by it, if it is not unity that is obtain x² + ba

x + ca = 0
Step-III: Shift the constant term ca on R.H.S. to get x² + b/a + x = c/a

Step-IV : Add square of half of the coefficient of x. i.e.  on both sides to obtain

Step-V : Write L.H.S. as the perfect square and simplify R.H.S. to get 

Step-VI : Take square root of both sides to get 

Step-VII : Obtain the values of x by shifting the constant term b/a on R.H.S. i.e. 

Ex.6 Solve : 9x² – 15x + 6 = 0

Sol. Here, 9x² – 15x + 6 = 0

 [Dividing throughout by 9]

 [Shifting the constant term on RHS)

[Adding square of half of coefficient of x on both sides]

 [Taking square root of both sides]

 x = 1 or, x = 2/3

Ex.7 Solve the equation  by the method of completing the square.

Sol. We have,

Hence, the roots are √3 and 1.

Solution by Quadratic Formula "Sreedharacharya's Rule"

Consider quadratic equation ax² + bx + c = 0, a  0 then  

 The roots of x are 

Thus, if D = b² – 4ac ≥ 0, then the quadratic equation ax² + bx + c = 0 has real roots α and β given by

Discriminant : If ax² + bx + c = 0, a  0 (a,b,c  R) is a quadratic equation, then the expression b² –4ac is known as its discriminant and is generally denoted by D or Δ.
 

Ex.8 Solve the quadratic equation x² – 6x + 4 = 0 by using quadratic formula (Sreedharacharya's Rule).

Sol. On comparing the given equation x² – 6x + 4 = 0 with the standard quadratic equation ax² + bx + c = 0,
we get a = 1, b = – 6, c = 4

Hence the required roots are

COMPETITION WINDOW
 SOLUTIONS OF EQUATIONS REDUCIBLE TO QUADRATIC FORM

Equations which are not quadratic at a glance but can be reduced to quadratic equations by suitable transformations.
Some of the common types are :
Type-I : ax⁴ + bx² + c = 0
This can be reduced to a quadratic equation by substituting x² = y i.e., ay² + by + c = 0

Type-II : a{p(x)}² + b.p (x) + c = 0 where p(x) is an expression in 'x'

Put p(x) = y, {p(x)}² = y² to get the quadratic equation ay² + by + c = 0.
e.g. Solve (x² + 3x)² – (x² + 3x) – 6 = 0, x  R
Putting x² + 3x = y, we get y² – y – 6 = 0
Solving, we get y = 3 or – 2
 x² + 3x = 3 or x² + 3x = – 2

Type-III : ap(x) + b p(x)  = c, 

where p(x) is an expression in x.
Put p(x) = y to obtain the quadratic equation ay² – cy + b = 0
e.g. Solve 
 = 

Type-IV : (i) a +   + c = 0  (ii) a b + c = 0
If the coefficient of b in the given equation contains  
- 2 and put
. In case the coefficient of b is  , then replace x² +  by   + 2 and put   = y.

e.g. Solve 9 9 – 52 = 0
Putting x + 1/x = y, we get : 9(y² – 2) – 9y – 52 = 0

Type-V : (x + a) (x + b) (x + c) (x + d) + k = 0, such that a + b = c + d.

Rewrite the equation in the form {(x + a) (x + b)} · {(x + c) (x + d)} + k = 0
Put x² + x(a + b) = x² + x(c + d) = y to obtain a quadratic equation in y i.e. (y + ab) (y + cd) = k.
e.g. Solve (x + 1) (x + 2) (x + 3) (x + 4) = 120

∵1 + 4 = 2 + 3,
we write the equation in the following form :
{(x + 1) (x + 4)} {x + 2) (x + 3)} = 120 ⇒ (x² + 5x + 4) (x² + 5x + 6) = 120
Putting x² + 5x = y, we get (y + 4) (y + 6) = 120
⇒ y = – 16 or 6 ⇒ x² + 5x = – 16 or x2 + 5x = 6
⇒ x = – 6 or 1  (x² + 5x + 16 has no real solution)

Type-VI :  = (cx + d) Square both sides to obtain (ax + b) = (cx + d)² 
or c²x² + (2cd – a) x + d² – b = 0
Reject those values of x, which do not satisfy both ax + b ≥ 0 and cx + d ≥ 0.
e.g. Solve :  + x = 1 3 ⇒ (2x + 9) = (13 – x)² (on squaring both sides)
⇒ x² – 28x + 160 = 0
⇒ x = 20 or 8 x = 20 does not satisfy 2x + 9 ≥ 0. So, x = 8 is the only root.

Type-VII :  = d x + e
Square both sides to obtain the quadratic equation x² (a – d²) + x (b – 2de) + (c – e²) = 0.
Solve it and reject those values of x which do not satisfy ax² + bx + c ≥ 0 and dx + e ≥ 0.

e.g. Solve  = x - 3 ⇒ 3x² + x + 5 = (x – 3)² (On squaring both sides)
⇒ 2x² + 7x – 4 = 0  ⇒ x = 1 2
 or – 4 No value of x satisfy 3x² + x + 5 ≥ 0 and x – 3 ≥ 0

Type-VIII := e Square both sides and simplify in such a manner that the expression involving radical sign on one side and all other terms are on the other side. Square both sides of the equation thus obtained and simplify it to obtain a quadratic in x. Reject these values which do not satisfy ax + b ≥ 0 and cx + d ≥ 0.

e.g. Solve : = 5

⇒ x = 0 or – 5
Clearly, x = 0 and x = – 5 satisfy 4 – x ≥ 0 and x + 9 ≥ 0.
Hence, the roots are 0 and – 5.