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## Equations Reducible to Quadratic Form

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:

Squaring on both sides, we have

 (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

### 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