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

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