What is the sum of all positive real solutions to the equation \frac{x^2 - 3x}{x^2 - 3x - … — SAT

What is the sum of all positive real solutions to the equation

\frac{x^2 - 3x}{x^2 - 3x - 2} + \frac{x^2 - 3x - 2}{x^2 - 3x} = \frac{5}{2}

Answer: 7

Answer

The sum of all positive real solutions is 7.

By substituting

u = x^2 - 3x

, the original rational equation simplifies to

\frac{u}{u - 2} + \frac{u - 2}{u} = \frac{5}{2}

. Multiplying both sides by the common denominator

2u(u-2)

and simplifying results in the quadratic equation

u^2 - 2u - 8 = 0

. Factoring gives

(u-4)(u+2) = 0

, so

u = 4

u = -2

. Substituting back

x^2 - 3x

for

u

leads to two quadratic equations:

x^2 - 3x = 4

(which has solutions

x = 4

and

x = -1

) and

x^2 - 3x = -2

(which has solutions

x = 2

and

x = 1

). Checking the denominators, none of these solutions make the original denominators zero, so they are all valid. The positive solutions are

1

2

, and

4

, and their sum is

1 + 2 + 4 = 7

Step-by-Step Solution

Introduce a substitution variable to simplify the rational equation.

Letting

u = x^2 - 3x

transforms the equation into

\frac{u}{u - 2} + \frac{u - 2}{u} = \frac{5}{2}

This reduces the degree of the rational expression and simplifies the algebraic manipulation required to solve it.

Eliminate the denominators by multiplying by the least common denominator.

Multiplying by

2u(u-2)

gives

2u^2 + 2(u-2)^2 = 5u(u-2)

, which simplifies to

u^2 - 2u - 8 = 0

This converts the rational equation into a standard quadratic equation in terms of

u

Solve the quadratic equation for

u

by factoring.

(u - 4)(u + 2) = 0

, which gives

u = 4

u = -2

Finding the values of

u

allows us to set up equations to solve for the original variable

x

Substitute back

x^2 - 3x

for

u

and solve the resulting quadratic equations for

x

From

x^2 - 3x = 4

, we get

(x-4)(x+1) = 0 \implies x = 4, -1

. From

x^2 - 3x = -2

, we get

(x-2)(x-1) = 0 \implies x = 2, 1

This yields all real values of

x

that satisfy the original algebraic structure.

Filter for positive real solutions and calculate their sum.

The positive solutions are

1

2

, and

4

. Their sum is

1 + 2 + 4 = 7

The question specifically asks for the sum of only the positive real solutions.

Key Concept

Solving rational equations using algebraic substitution and factoring quadratic equations.

Alternative Method

Instead of using substitution directly, the equation can be solved by multiplying by the common denominator

(x^2 - 3x - 2)(x^2 - 3x)

to get a fourth-degree polynomial:

2(x^2 - 3x)^2 + 2(x^2 - 3x - 2)^2 = 5(x^2 - 3x)(x^2 - 3x - 2)

. Letting

z = x^2 - 3x

at this stage simplifies this expression to

2z^2 + 2(z-2)^2 = 5z(z-2)

, which avoids full expansion into a fourth-degree polynomial and leads to the same quadratic in

z

Estimated Time:3m 0s

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Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

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