Examkin

Soru

Zorluk: Çok zorRadical and Rational Equations
What is the sum of all positive real solutions to the equation
x2−3xx2−3x−2+x2−3x−2x2−3x=52\frac{x^2 - 3x}{x^2 - 3x - 2} + \frac{x^2 - 3x - 2}{x^2 - 3x} = \frac{5}{2}
?

Cevap: 7

Cevap

The sum of all positive real solutions is 7.
By substituting u=x2−3xu = x^2 - 3x, the original rational equation simplifies to uu−2+u−2u=52\frac{u}{u - 2} + \frac{u - 2}{u} = \frac{5}{2}. Multiplying both sides by the common denominator 2u(u−2)2u(u-2) and simplifying results in the quadratic equation u2−2u−8=0u^2 - 2u - 8 = 0. Factoring gives (u−4)(u+2)=0(u-4)(u+2) = 0, so u=4u = 4 or u=−2u = -2. Substituting back x2−3xx^2 - 3x for uu leads to two quadratic equations: x2−3x=4x^2 - 3x = 4 (which has solutions x=4x = 4 and x=−1x = -1) and x2−3x=−2x^2 - 3x = -2 (which has solutions x=2x = 2 and x=1x = 1). Checking the denominators, none of these solutions make the original denominators zero, so they are all valid. The positive solutions are 11, 22, and 44, and their sum is 1+2+4=71 + 2 + 4 = 7.

Adım Adım Çözüm

1
Introduce a substitution variable to simplify the rational equation.
Letting u=x2−3xu = x^2 - 3x transforms the equation into uu−2+u−2u=52\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.
2
Eliminate the denominators by multiplying by the least common denominator.
Multiplying by 2u(u−2)2u(u-2) gives 2u2+2(u−2)2=5u(u−2)2u^2 + 2(u-2)^2 = 5u(u-2), which simplifies to u2−2u−8=0u^2 - 2u - 8 = 0.
This converts the rational equation into a standard quadratic equation in terms of uu.
3
Solve the quadratic equation for uu by factoring.
(u−4)(u+2)=0(u - 4)(u + 2) = 0, which gives u=4u = 4 or u=−2u = -2.
Finding the values of uu allows us to set up equations to solve for the original variable xx.
4
Substitute back x2−3xx^2 - 3x for uu and solve the resulting quadratic equations for xx.
From x2−3x=4x^2 - 3x = 4, we get (x−4)(x+1)=0  ⟹  x=4,−1(x-4)(x+1) = 0 \implies x = 4, -1. From x2−3x=−2x^2 - 3x = -2, we get (x−2)(x−1)=0  ⟹  x=2,1(x-2)(x-1) = 0 \implies x = 2, 1.
This yields all real values of xx that satisfy the original algebraic structure.
5
Filter for positive real solutions and calculate their sum.
The positive solutions are 11, 22, and 44. Their sum is 1+2+4=71 + 2 + 4 = 7.
The question specifically asks for the sum of only the positive real solutions.

Anahtar Kavram

Solving rational equations using algebraic substitution and factoring quadratic equations.

Alternatif Yöntem

Instead of using substitution directly, the equation can be solved by multiplying by the common denominator (x2−3x−2)(x2−3x)(x^2 - 3x - 2)(x^2 - 3x) to get a fourth-degree polynomial: 2(x2−3x)2+2(x2−3x−2)2=5(x2−3x)(x2−3x−2)2(x^2 - 3x)^2 + 2(x^2 - 3x - 2)^2 = 5(x^2 - 3x)(x^2 - 3x - 2). Letting z=x2−3xz = x^2 - 3x at this stage simplifies this expression to 2z2+2(z−2)2=5z(z−2)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 zz.
Tahmini Süre:3m 0s
Bu soruyu puanla

Konular

Radical and Rational Equations

Bu konudan daha fazla soru