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Example 1: Solve the equation
x - 3 sqrt ( x ) = - 2
Solution to Example 1:
- Let u = sqrt ( x ) so that u 2 = x. We now substitute x and sqrt (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 sqrt ( x ) and solve for x
sqrt ( x ) = 1 or sqrt ( x ) = 2
x = 1 or x = 4
Example 2: Use the method of sustitution 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 sustitution 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 sustitution 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 sustitution to solve the equation
sin 2 x - 4 sin x - 5 = 0
for x in the interval [ 0 , 2 Pi ). 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 Pi / 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.
More references and links on how to Solve Equations, Systems of Equations and Inequalities.
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