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Difficulty: Very hardRadical and Rational Equations

What is the sum of all real solutions to the equation 3x+7−103x+7=3\sqrt{3x + 7} - \frac{10}{\sqrt{3x + 7}} = 3?

Answer: 6

Answer

The only real solution is 6, so the sum of all real solutions is 6.
The correct answer is 6. By substituting u=3x+7u = \sqrt{3x + 7}, the equation becomes u−10/u=3u - 10/u = 3, which simplifies to the quadratic u2−3u−10=0u^2 - 3u - 10 = 0. Factoring gives (u−5)(u+2)=0(u - 5)(u + 2) = 0. Since the principal square root must be non-negative, uu must be positive, so we reject u=−2u = -2 and keep u=5u = 5. Solving 3x+7=5\sqrt{3x + 7} = 5 by squaring both sides yields 3x+7=253x + 7 = 25, which simplifies to x=6x = 6. The extraneous solution x=−1x = -1 (which comes from u=−2u = -2) must be discarded because substituting it back into the original equation results in −3-3 instead of 33. Thus, the only real solution is 6.

Step-by-Step Solution

1
Substitute u=3x+7u = \sqrt{3x + 7} into the equation.
u−10u=3u - \frac{10}{u} = 3, where u>0u > 0.
This substitution simplifies the radical equation into a rational equation of a single variable.
2
Multiply the entire equation by uu and solve the resulting quadratic equation.
u2−3u−10=0  ⟹  (u−5)(u+2)=0u^2 - 3u - 10 = 0 \implies (u - 5)(u + 2) = 0, which yields u=5u = 5 or u=−2u = -2.
Clearing the denominator transforms the rational expression into a standard quadratic equation.
3
Apply the constraint that u=3x+7u = \sqrt{3x + 7} must be non-negative.
u=5u = 5 is the only valid root; u=−2u = -2 has no real solution and must be discarded.
The output of a principal square root is non-negative by definition.
4
Substitute back 3x+7=5\sqrt{3x + 7} = 5 and solve for xx.
3x+7=25  ⟹  3x=18  ⟹  x=63x + 7 = 25 \implies 3x = 18 \implies x = 6.
Squaring both sides eliminates the radical and allows us to isolate the variable xx.
5
Verify if solving 3x+7=−2\sqrt{3x+7} = -2 leads to an extraneous solution.
Squaring both sides of 3x+7=−2\sqrt{3x+7} = -2 gives 3x+7=4  ⟹  x=−13x+7 = 4 \implies x = -1. Substituting x=−1x = -1 into the original equation yields 4−104=2−5=−3≠3\sqrt{4} - \frac{10}{\sqrt{4}} = 2 - 5 = -3 \neq 3. Therefore, x=−1x = -1 is extraneous.
Verifying solutions is necessary to identify and discard values that do not satisfy the original equation.

Key Concept

Solving equations using substitution and identifying extraneous solutions
Estimated Time:2m 30s
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Topics

Radical and Rational Equations

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