Some equations can be reduced to the quadratic form by using proper substitution. Let’s now solve such problems.

Example 1

Solve ( x² - x )² - 8( x² - x ) + 12 = 0

Solution:

Put x² - x = a in the given equation.

Then a² - 8a + 12 = 0

a² - 8a + 12 = a² - 6a - 2a + 12

                   = a( a - 6) - 2( a - 6 )

                   = ( a - 2 ) ( a - 6 )

i.e., a² - 8a + 12 = 0 is

( a - 2 ) ( a - 6 ) = 0

Therefore a = 2 or a = 6

1. If a = 2, we have x² - x = 2
   i.e., x² - x - 2 = 0
   x² - 2x + x - 2 = 0
   x ( x - 2 ) + 1 ( x - 2 ) = 0
   ( x + 1 ) ( x - 2 ) = 0 from which
   we get x = -1 or 2
2. If a = 6, we have x² - x - 6 = 0
   x² - 3x + 2x - 6 = 0
   x ( x - 3 ) + 2 ( x - 3 ) = 0
   ( x + 2 ) ( x - 3 ) = 0
   i.e., x = - 2 or 3
   Therefore, the roots of the given equation are -1, -2, 2, 3
   i.e., Solution set = {- 1, - 2, 2, 3}

Example 2

Solve + = 0

Solution:

(x + 1) + (2x + 3) + 2 = 0

3x - 21 = -2

3x - 21 = -2

Once again squaring,

( 3x - 21 )² = 4( x + 1 ) ( 2x + 3 )

9x² + 441 - 126x = 8x² + 20x + 12

x² - 146x + 429 = 0

x² - 143x - 3x + 429 = 0

x ( x - 143 ) - 3 ( x - 143 ) = 0

( x - 3 ) ( x - 143 )

Therefore, x = 3 or x = 143.

Clearly, x = 143 does not satisfy the given equation.

x = 3 satisfies the equation.

Therefore x = 3 is the only solution of the given equation.

Example 3

Solve (x² + 2x)² - 11 (x² + 2x) + 24 = 0

(x² + 2x)² - 11 (x² + 2x) + 24 = 0

Put x² + 2x = a      ---------------------- (1)

Then the equation becomes a² - 11a + 24 = 0

a² - 11a + 24 = 0 = a² - 8a - 3a + 24

a(a - 8) -3 (a - 8) = 0

(a - 8) (a - 3) = 0

a - 8 = 0 (or) a - 3 = 0

a = 8 (or) a = 3

When a = 8, equation (1) becomes x² + 2x = 8

That is, x² + 2x - 8 = 0

x² + 4x - 2x - 8 = 0

x(x + 4) - 2(x + 4) = 0

(x + 4) (x - 2) = 0

x = -4 or 2

When a = 3, equation (1) becomes x² + 2x = 3

That is, x² + 2x - 3 = 0

x² + 3x - x - 3 = 0

x(x + 3) -1(x + 3) = 0

x(x + 3)(x - 1) = 0

x = -3 or 1

Therefore, x = 1, 2, -3 or -4

Try these questions

Solve the following equations

1. x⁴ - 5x² + 4 = 0
2. x⁶ - 9x³+ 8 = 0
3. ( x² - 2x )² -- 11 ( x² -- 2x ) + 24 = 0
4. 4x⁴-- 17x² + 4 = 0
5. 9x⁴ + 25 = 30x²
6. ( 2x² -- 3x )² -- 4 ( 2x²-- 3x ) -- 5 = 0
7. ( x² + 5x + 3 ) + 3 / ( x² + 5x + 7 ) = 0
8. 9 / ( 1 + x + x² ) = 5 -- x -- x²
9. x ( x + 1 ) ( x + 2 ) ( x + 3 ) = 120
10. ( x + 1 ) ( x + 3 ) ( x + 4 ) ( x + 6 ) = 280

Answers

1. x⁴ -- 5x² + 4 = 0
   (x²)²-- 5 (x²) + 4 = 0
   Put x² = a
   Then the equation becomes
   a²-- 5a + 4 =0
   a²-- 4a -- a + 4 = 0
   a ( a -- 4 ) --1 ( a -- 4 ) = 0
   ( a -- 4 ) ( a -- 1 ) = 0
   a -- 4 = 0 or a -- 1 = 0
   a = 4 or a = 1
   If a = 4 then x² = 4
   x = √4= ± 2
   If a = 1 then x² = 1
   x = √1= ± 1
   Therefore x = 2 or + 1, -1, -2


2. 2-- 9x³+ 8 = 0
   ( x³ )² -- 9( x³ ) + 8 = 0
   Put x³ = a
   Then the equation becomes
   a²-- 9a + 8 = 0
   a²-- 8a -- a + 8 = 0
   a( a -- 8 ) --1( a -- 8 ) = 0
   ( a -- 8 ) ( a -- 1 ) = 0
   a -- 8 =0 or a -- 1 = 0
   a = 8 or a = 1
   If a = 8, then x³ = 8
   x³ = 2³
   x = 2
   If a = 1, then x³ = 1
   x³ = 1³
   x = 1
   Therefore x = 2 or 1


3. ( x² -- 2x )² -- 11 ( x² -- 2x ) + 24 = 0
   Put x² -- 2x = a
   Then the equation becomes a²-- 11a + 24 =0
   That is a²-- 8a -- 3a + 24 = 0
   A ( a -- 8 ) -- 3 ( a -- 8 ) = 0
   ( a -- 8 ) ( a -- 3 ) = 0
   a -- 8 = 0 or a -- 3 = 0
   a = 8 or a = 3
   If a = 8 then x² -- 2x = a will change to x² - 2x = 8
   That is x² -- 2x -- 8 = 0
   x²-- 4x + 2x -- 8 = 0
   x ( x -- 4 ) + 2( x -- 4 ) = 0
   ( x -- 4 ) ( x + 2 ) = 0
   x -- 4 = 0 or x + 2 = 0
   x = 4 or x = -- 2
   If a = 3 then x² -- 2x = a will change as x² - 2x = 3
   That is x² -- 2x -- 3 = 0
   x² -- 3x + x -- 3 = 0
   x ( x -- 3 ) + 1( x -- 3 ) = 0
   ( x -- 3 ) ( x + 1 ) = 0
   x -- 3 = 0 or x + 1 = 0
   x = 3 or x = -- 1
   Therefore x = -- 1, -- 2, 3 or 4


4. 4x⁴ -- 17x² + 4 = 0
   4( x² )² -- 17( x² ) + 4 = 0
   Put x² = a
   Then the equation becomes 4a² -- 17a + 4 = 0
   4a²-- 16a -- a + 4 = 0
   4a ( a -- 4 ) -- 1 ( a -- 4 ) = 0
   ( a -- 4 ) ( 4a -- 1 ) = 0
   a -- 4 = 0 or 4a -- 1 = 0
   a = 4; or 4a = 1 or a = ¼
   When a = 4 then x² = a changes to x² = 4
   Therefore x = √4 = ± 2
   a = 1/4 then x² = a changes to x² = 1/4z
   Therefore x = √1/4= ± 1/2
   x = + 2 or + 1/2, -2, -1/2


5. 9x⁴+ 25 = 30x²
   That is 9x⁴ -- 30x² + 25 = 0
   ( 3x² )² -- 2 * 3x² * 5 + (5)² = 0
   This is of the form a² -- 2ab + b² = ( a -- b )²
   Therefore ( 3x² ) -- 2* 3x² * 5 + (5)² = 0 = ( 3x² -- 5 )²
   ( 3x² -- 5 )² = 0
   3x²-- 5 = 0
   That is, 3x² = 5 (or) x² = 5/3.
   Therefore x = √5/3 = ± 5/3


6. ( 2x² -- 3x )² -- 4 ( 2x²-- 3x ) -- 5 = 0
   Put 2x²-- 3x = a
   Then the equation becomes a²-- 4a -- 5 = 0
   a²-- 4a -- 5 = 0
   a² -- 5a + a -- 5 = 0
   a ( a -- 5 ) + 1 ( a -- 5 ) = 0
   ( a -- 5 ) ( a + 1 ) = 0
   a -- 5 = 0 or a + 1 = 0
   a = 5 or a = -1
   When a = 5
   Therefore 2x²-- 3x = 5
   or 2x² -- 3x -- 5 = 0
   2x² -- 5x + 2x -- 5 = 0
   x ( 2x -- 5 ) + 1( 2x -- 5 ) = 0
   ( 2x -- 5 ) ( x + 1 ) = 0
   2x -- 5 = 0 or x + 1 = 0
   2x -- 5 = 0           x = -1
   2x = 5
   x = 5/2
   Again when a = -- 1 then 2x² -- 3x = -- 1
   That is 2x²-- 3x + 1 = 0
   2x² -- 2x -- x + 1 = 0
   2x ( x -- 1 ) -- 1 ( x -- 1 ) = 0
   ( x -- 1 ) ( 2x -- 1 ) = 0
   x -- 1 = 0 or 2x -- 1 = 0
   x -- 1 = 0 or 2x = 1
   x = 1       or x = 1/2
   Therefore x = + 1,1/2 or 5/2, -1


7. (x²+ 5x + 3) + 3 / (x²+ 5x +7) = 0
   Put x² + 5x = a
   Then the equation becomes (a + 3) + 3 / (a + 7) = 0
   (a + 3) (a + 7) + 3 / (a + 7) = 0
   a² + 3a + 7a + 21 + 3 = 0
   a²+ 10a + 21 + 3 = 0
   a² + 10a + 24 = 0
   a² + 6a + 4a + 24 = 0
   a ( a + 6 ) + 4 ( a + 6 ) = 0
   ( a + 6 ) ( a + 4 ) = 0
   a + 6 = 0 or a + 4 = 0
   a = -- 6 or a = -- 4
   When a = -- 6
   Then x²+ 5x = -- 6
   x²+ 5x + 6 = 0
   x² + 3x + 2x + 6 = 0
   x ( x+ 3 ) + 2( x + 3 ) = 0
   ( x + 3 ) ( x + 2 ) = 0
   x + 3 = 0 or x + 2 = 0
   x = -- 3 or x = -- 2
   When a = -- 4, then x² - 5x = --4
   x²+ 5x + 4 = 0
   x² + 4x + x + 4 = 0
   x ( x + 4 ) + 1 ( x + 4 ) = 0
   ( x + 4 ) ( x + 1 ) = 0
   x + 4 =0 or x + 1 = 0
   x = -- 4 or x = -- 1
   Therefore x =-1, -- 2, -- 3 or -- 4.


8. 9/ (1 + x + x²) = 5 -- x -- x²
   9/ (1 + x + x²) = 5 -- ( x + x² )
   Put x + x² = a
   Then the equation becomes 9/ (1 + a) = 5 -- a
   Cross multiplying 9 = 5 + 5a -- a -- a² (or) 9 -- 5 -- 5a + a + a²= 0
   4 - 4a + a² = 0
   a² -- 4a + 4 = 0
   (a)² -- 2a * 2 + (2)² = 0
   ( a -- 2 )² = 0
   [∴ This of the form a² - 2ab + b2 = (a - b)²]
   a -- 2 = 0
   When a = 2
   Then x + x² = 2 or x² + x -- 2 = 0
   x²+ 2x -- x -- 2 = 0
   x ( x + 2 ) -- 1 ( x + 2 ) = 0
   ( x + 2 ) ( x -- 1 ) = 0
   x + 2 = 0 or x -- 1 = 0
   x = -- 2 or x = 1
   x = 1 or -- 2


9. x ( x + 1 ) ( x + 2 ) ( x + 3 ) = 120
   Regrouping the factors:
   [ x ( x + 3 ) ] [ ( x + 1 ) ( x + 2 ) ] = 120
   ( x²+ 3x ) ( x²+ 3x + 2 ) = 120
   Putting x²+ 3x = a, the equation becomes
   a ( a + 2 ) = 120
   a²+ 2a = 120 (or) a²+ 2a -- 120 = 0
   a² + 12a -- 10a -- 120 = 0
   a ( a +12 ) -- 10 ( a+ 12 ) = 0
   (a + 12) (a -- 10) = 0
   a + 12 = 0 or a -- 10 = 0
   a = -- 12 or a = 10
   If a = -- 12 then x²+ 3x = 12
   x²+ 3x + 12 = 0
   (12 has no real factors such that their sum is 3. So no real roots)
   If a = 10 then x²+ 3x = 10
   x² + 3x -- 10 = 0
   That is 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


10. ( x + 1 ) ( x + 3 ) ( x + 4 ) ( x + 6 ) = 280
    [ ( x + 1 ) ( x + 6 ) ] [( x + 3 ) ( x + 4 ) ] = 280
    ( x²+ 7x + 6 ) ( x²+ 7x + 12 ) = 280
    Put x² + 7x + 6 = a
    Then the equation becomes a( a + 6 ) = 280
    a²+ 6a = 280
    a² + 6a -- 280 = 0
    -- 280 = 20 * ( -- 14); 20 + ( -- 14) = 6
    a²+ 20a -- 14a -- 280 = 0
    a( a + 20 ) -- 14( a+ 20 ) = 0
    ( a + 20 ) ( a -- 14 ) = 0
    a = -- 20 or 14
    If a = -- 20 then
    x² + 7x + 6 = -20
    x² + 7x + 6 + 20 = 0
    x²+ 7x -- 26 = 0
    (14 has no real numbers as factors whose different is 7; hence no real roots)
    If a = 14 then x²+ 7x + 6 = 14
    x² + 7x + 6 -- 14 = 0
    x² + 7x -- 8 = 0
    That is x² + 8x -- x -- 8 = 0
    x ( x + 8 ) -- 1 ( x + 8 ) = 0
    ( x + 8 ) ( x -- 1 ) = 0
    x + 8 = 0 or x -- 1 = 0
    x = -- 8 or 1