Let f(x)=(x^2+5x)(x^2+5x+4). Solve √(f(x)+4)=2501 and find the sum of all real roots.
Answer and solution
Correct answer: B
Solution
Answer: B — -5
1. Let u=x²+5x. Then f(x)+4=u(u+4)+4=(u+2)².
3. For u=2499, x²+5x−2499=0 has discriminant 10021>0, hence two real roots.
4. For u=−2503, x²+5x+2503=0 has discriminant 25−10012<0, hence no real roots.
5. By Vieta, the two valid roots sum to −5, option B. Do not include the complex-root branch.
Why this method works:
The square root of a square is an absolute value, not automatically the original expression. Therefore both signs of u+2 must be examined after completing the square; ignoring the negative branch without explanation is incomplete.
Independent check and common trap:
Complete x²+5x=(x+5/2)²−25/4. Hence u cannot equal−2503, which is below its minimum−6.25. The remaining quadratic has two real roots and their sum−5 by Vieta, avoiding calculation of the large radicals.
Check: Complete x²+5x=(x+5/2)²−25/4. Hence u cannot equal−2503, which is below its minimum−6.25. The remaining quadratic has two real roots and their sum−5 by Vieta, avoiding calculation of the large radicals.
1. Let u=x²+5x. Then f(x)+4=u(u+4)+4=(u+2)².
| 2. The nonnegative square root is | u+2 | , so | u+2 | =2501 gives u=2499 or u=−2503. |
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3. For u=2499, x²+5x−2499=0 has discriminant 10021>0, hence two real roots.
4. For u=−2503, x²+5x+2503=0 has discriminant 25−10012<0, hence no real roots.
5. By Vieta, the two valid roots sum to −5, option B. Do not include the complex-root branch.
Why this method works:
The square root of a square is an absolute value, not automatically the original expression. Therefore both signs of u+2 must be examined after completing the square; ignoring the negative branch without explanation is incomplete.
Independent check and common trap:
Complete x²+5x=(x+5/2)²−25/4. Hence u cannot equal−2503, which is below its minimum−6.25. The remaining quadratic has two real roots and their sum−5 by Vieta, avoiding calculation of the large radicals.
G Strategy shortcut
Spot: Quadratic Equations and Perfect Squares.| Apply: Set u=x²+5x. The radical becomes | u+2 | . Of u=2499 and −2503, only the first is possible since x²+5x≥−25/4. Vieta gives −5. |
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G Strategy: Spot: Quadratic Equations and Perfect Squares. Apply: Set u=x²+5x. The radical becomes |u+2|. Of u=2499 and −2503, only the first is possible since x²+5x≥−25/4. Vieta gives −5. Check: Complete x²+5x=(x+5/2)²−25/4. Hence u cannot equal−2503, which is below its minimum−6.25. The remaining quadratic has two real roots and their sum−5 by Vieta, avoiding calculation of the large radicals.