| The number of integer solutions of 3 | x | (x^2+2)=7x^2 is ______. |
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Answer and solution
Correct answer: 1
Solution
Answer: 1
2. Factor: t(3t²−7t+6)=0. The t=0 case gives x=0.
3. For t≠0, solve 3t²−7t+6=0. Its discriminant is 49−72=−23, so there are no real t.
4. Thus the original equation has only the real solution x=0, which is an integer. Count=1.
Why this method works:
Independent check and common trap:
The quadratic 3t²−7t+6 has positive leading coefficient and negative discriminant, so it is strictly positive for all realt. Therefore only t=0 makes the product zero. Substitutionx=0 gives 0=0, and the integer-solution count is 1.
Check: The quadratic 3t²−7t+6 has positive leading coefficient and negative discriminant, so it is strictly positive for all realt. Therefore only t=0 makes the product zero. Substitutionx=0 gives 0=0, and the integer-solution count is 1.
| 1. Set t= | x | ≥0, so x²=t². Equation becomes 3t(t²+2)=7t². |
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2. Factor: t(3t²−7t+6)=0. The t=0 case gives x=0.
3. For t≠0, solve 3t²−7t+6=0. Its discriminant is 49−72=−23, so there are no real t.
4. Thus the original equation has only the real solution x=0, which is an integer. Count=1.
Why this method works:
| Replacing | x | by t does not discard negative x because x²=t² for both signs. Factor outt before dividing; otherwise the valid zero solution would be lost. A positive t would correspond to two opposite x values, but none exists here. |
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Independent check and common trap:
The quadratic 3t²−7t+6 has positive leading coefficient and negative discriminant, so it is strictly positive for all realt. Therefore only t=0 makes the product zero. Substitutionx=0 gives 0=0, and the integer-solution count is 1.
G Strategy shortcut
Spot: Absolute Value and Quadratic Equations.| Apply: Put t= | x | . Then t[3t²−7t+6]=0. The quadratic has Δ=−23, so only t=0 and x=0; answer 1. |
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G Strategy: Spot: Absolute Value and Quadratic Equations. Apply: Put t=|x|. Then t[3t²−7t+6]=0. The quadratic has Δ=−23, so only t=0 and x=0; answer 1. Check: The quadratic 3t²−7t+6 has positive leading coefficient and negative discriminant, so it is strictly positive for all realt. Therefore only t=0 makes the product zero. Substitutionx=0 gives 0=0, and the integer-solution count is 1.