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The powerful method of substitution is used to solve different types of equations.

Example 1

Solve the equation
x - 3 √x = - 2

Solution to Example 1

• Let u = √x) so that u 2 = x. We now substitute x and √ (x) by u and u 2 respectively to obtain an equation in u.
u 2 - 3 u = - 2
• The above is a quadratic equation; rewrite it so that its right term is equal to zero.
u 2 - 3 u + 2 = 0
• Use any method to solve the above equation for u to obtain:
u = 1 or u = 2
• We now substitute u by √x and solve for x
√x = 1 or √x = 2
x = 1 or x = 4

Example 2

Use the method of substitution to solve the equation
1 / (x - 1) 2 - 1 / (x - 1) - 2 = 0

Solution to Example 2

• Let u = 1 / (x - 1) and make the substitution in the above equation to obtain an equation in u.
u 2 - u - 2 = 0
• Solve the above quadratic equation to obtain:
u = - 1 or u = 2
• We now substitute u by 1 / (x - 1) and solve for x
1 / (x - 1) = - 1 or 1 / (x - 1) = 2
x = 0 or x = 3 / 2

Example 3:

Use the method of substitution to solve the equation
- (x + 3) 6 + 4 (x + 3) 3 = - 21

Solution to Example 3

• Let u = (x + 3) 3 and substitute in the above equation to obtain an equation in u.
- u 2 + 4 u = -21
u 2 - 4 u - 21 = 0
• Solve the above quadratic equation to obtain:
u = - 3 or u = 7
• We now substitute u by (x + 3) 3 and solve for x
(x + 3) 3 = - 3 or (x + 3) 3 = 7
• We now solve for x
x = - 3 - cube root ( 3 )
or
x = - 3 + cube root ( 7 )

Example 4

Use the method of substitution to solve the equation
3 e 2 x - e x - 2 = 0

Solution to Example 4

• Let u = e x so that u 2 = e 2 x and substitute in the above equation to obtain:
3 u 2 - u - 2 = 0
• Use any method to solve the above quadratic equation.
u = 1
or
u = - 2 / 3
• We now substitute u by e x and solve for x
e x = 1 or e x = - 2 / 3
• We now solve e x = 1 to obtain:
x = 0
• The second equation e x = - 2 / 3 has no solution since e x is always positive.

Example 5

Use the method of substitution to solve the equation
sin 2 x - 4 sin x - 5 = 0
for x in the interval [ 0 , 2 π ). Give x in radians.

Solution to Example 5

• Let u = sin x and substitute in the above equation to obtain:
u 2 - 4 u - 5 = 0
• Solve the above quadratic equation.
u = - 1
or
u = 5
• We now substitute u by sin x and solve for x
sin x = - 1 or sin x = 5
• We solve sin x = - 1 to obtain:
x = 3 π / 2
• The range of sin x is the set of values in the interval [ - 1 , 1 ] and therefore the equation sin x = 5 has no solution.

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