Solving an Equation with Rational Exponents (2.9.4) Flashcards
• To solve an equation containing a single rational exponent, isolate the expression with the rational exponent on one side of the equation and then raise each side to the reciprocal of that rational exponent.
• To solve an equation containing a single rational exponent, isolate the expression with the rational exponent on one side of the equation and then raise each side to the reciprocal of that rational exponent.
• To solve an equation containing two rational exponents (one on each side), raise each side to the LCM (least common multiple) of the two rational exponents’ denominators. For example, if the left side is an expression raised to the 1/4 power and the right side is an expression raised to the 1/2 power, then raise both sides to the 4th power since 4 is the LCM of the two rational exponents’ denominators, 2 and 4.
• To solve an equation containing two rational exponents (one on each side), raise each side to the LCM (least common multiple) of the two rational exponents’ denominators. For example, if the left side is an expression raised to the 1/4 power and the right side is an expression raised to the 1/2 power, then raise both sides to the 4th power since 4 is the LCM of the two rational exponents’ denominators, 2 and 4.
• Extraneous solutions do exist for these equations. Always check the possible solutions to determine which, if any, are solutions for the equation.
• Extraneous solutions do exist for these equations. Always check the possible solutions to determine which, if any, are solutions for the equation.
Notice that this equation contains two rational exponents.
The left side is an expression raised to the 1/2 power and the
right side is also an expression raised to the 1/2 power. So,
begin by raising each side to the 2nd power (since 2 is the
LCM of 2 and 2).
Raising each side to the 2nd power, i.e., squaring each side,
eliminates the rational exponents. Furthermore, neither side
includes an exponent at all since the product of 1/2 and 2 is
1. The resulting equation is a linear equation where the
variable is on both sides. So, solve by isolating x.
Be sure to check the solution, x = 4, to see if it is extraneous.
When 4 is substituted into the original equation for x, the
result is a true statement. Therefore, x = 4 is not extraneous.
The equation given here contains one rational exponent. The
power with the rational exponent is already isolated on one
side of the equation. Therefore, begin by raising both sides
to the 2nd power (since 2 is the reciprocal of 1/2).
In this example, the exponents do not completely cancel out.
The right side is left with a power of 2. So, simplify the right
side, (x + 1)2
, by using FOIL. The resulting equation is
quadratic. So, simplify to get 0 on one side, then factor and
apply the Zero-Factor Property. The two possible solutions
are x = −5 and x = 2.
Check for extraneous solutions. Substitute 2 into the original
equation for x and simplify. The result is 3 = 3, which is a
true statement. Therefore, x = 2 is a solution.
Substitute −5 into the original equation for x and simplify.
The result is 4 = −4, which is a false statement. Therefore,
x = −5 is a not a solution. Note that the square root of 16 is
equal to 4, not ±4. When the square root of a number is found,
only the principal, i.e. positive, square root is considered.
(However, when the square root is applied to both sides of
an equation, both the positive and the negative roots are
considered.)
Notice that this equation contains two rational exponents.
The left side is an expression raised to the 1/2 power and the
right side is also an expression raised to the 1/2 power. So,
begin by raising each side to the 2nd power (since 2 is the
LCM of 2 and 2).
Raising each side to the 2nd power, i.e., squaring each side,
eliminates the rational exponents. Furthermore, neither side
includes an exponent at all since the product of 1/2 and 2 is
1. The resulting equation is a linear equation where the
variable is on both sides. So, solve by isolating x.
Be sure to check the solution, x = 4, to see if it is extraneous.
When 4 is substituted into the original equation for x, the
result is a true statement. Therefore, x = 4 is not extraneous.
The equation given here contains one rational exponent. The
power with the rational exponent is already isolated on one
side of the equation. Therefore, begin by raising both sides
to the 2nd power (since 2 is the reciprocal of 1/2).
In this example, the exponents do not completely cancel out.
The right side is left with a power of 2. So, simplify the right
side, (x + 1)2
, by using FOIL. The resulting equation is
quadratic. So, simplify to get 0 on one side, then factor and
apply the Zero-Factor Property. The two possible solutions
are x = −5 and x = 2.
Check for extraneous solutions. Substitute 2 into the original
equation for x and simplify. The result is 3 = 3, which is a
true statement. Therefore, x = 2 is a solution.
Substitute −5 into the original equation for x and simplify.
The result is 4 = −4, which is a false statement. Therefore,
x = −5 is a not a solution. Note that the square root of 16 is
equal to 4, not ±4. When the square root of a number is found,
only the principal, i.e. positive, square root is considered.
(However, when the square root is applied to both sides of
an equation, both the positive and the negative roots are
considered.)
This equation contains two rational exponents. So, raise each
side to the power that is the LCM of the denominators. The
LCM of 3 and 3 is just 3, so cube each side.
Simplifying the exponents reveals that the left side is squared
(since the product of 2/3 and 3 is 2) and the right side no
longer contains an exponent (since the product of 1/3 and 3
is 1). Use FOIL to expand the right side, then subtract x from
each side to get 0 on one side. Next, factor and apply the
Zero-Factor Property. The two possible solutions are x = 1/4
and x = 1.
Substitute each possible solution into the original equation
and simplify to determine if either solution is extraneous.
Both substitutions result in true statements. Therefore, neither
solution is extraneous.
This equation contains two rational exponents. So, raise each
side to the power that is the LCM of the denominators. The
LCM of 3 and 3 is just 3, so cube each side.
Simplifying the exponents reveals that the left side is squared
(since the product of 2/3 and 3 is 2) and the right side no
longer contains an exponent (since the product of 1/3 and 3
is 1). Use FOIL to expand the right side, then subtract x from
each side to get 0 on one side. Next, factor and apply the
Zero-Factor Property. The two possible solutions are x = 1/4
and x = 1.
Substitute each possible solution into the original equation
and simplify to determine if either solution is extraneous.
Both substitutions result in true statements. Therefore, neither
solution is extraneous.
