Every Algebra Question Ever

Algebra Level pending

Solve for x x (a real root):

5 x 2 + 6 x + 5 = 64 x 3 + 4 x 5 x 2 + 6 x + 6 \sqrt{5x^{2}+6x+5} = \frac{64x^{3}+4x}{5x^{2}+6x+6}

Source: A Vietnamese Book


The answer is 1.

Bryan Mw by Bryan MW , 17, Malaysia

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1 solution

{ { x → 1 } , { x → − 1 361 1 3 ( 361 − 964637 + 36 i 136255611 3 + 3734545 − 964637 + 36 i 136255611 3 − 32981 ) − 1 2 − 4 − 964637 + 36 i 136255611 3 1083 − 263848 390963 − 41380 1083 − 964637 + 36 i 136255611 3 + 4000800 130321 1 3 ( 361 − 964637 + 36 i 136255611 3 + 3734545 − 964637 + 36 i 136255611 3 − 32981 ) − 39 361 } , { x → − 39 361 − 1 361 1 3 ( 361 − 964637 + 36 i 136255611 3 + 3734545 − 964637 + 36 i 136255611 3 − 32981 ) + 1 2 √ ( − 4 − 964637 + 36 i 136255611 3 1083 − 263848 390963 − 41380 1083 − 964637 + 36 i 136255611 3 + 4000800 130321 1 3 ( 361 − 964637 + 36 i 136255611 3 + 3734545 − 964637 + 36 i 136255611 3 − 32981 ) \left\{\{x\to 1\},\left\{x\to -\frac{1}{361} \sqrt{\frac{1}{3} \left(361 \sqrt[3]{-964637+36 i \sqrt{136255611}}+\frac{3734545}{\sqrt[3]{-964637+36 i \sqrt{136255611}}}-32981\right)}-\frac{1}{2} \sqrt{-\frac{4 \sqrt[3]{-964637+36 i \sqrt{136255611}}}{1083}-\frac{263848}{390963}-\frac{41380}{1083 \sqrt[3]{-964637+36 i \sqrt{136255611}}}+\frac{4000800}{130321 \sqrt{\frac{1}{3} \left(361 \sqrt[3]{-964637+36 i \sqrt{136255611}}+\frac{3734545}{\sqrt[3]{-964637+36 i \sqrt{136255611}}}-32981\right)}}}-\frac{39}{361}\right\},\left\{x\to -\frac{39}{361}-\frac{1}{361} \sqrt{\frac{1}{3} \left(361 \sqrt[3]{-964637+36 i \sqrt{136255611}}+\frac{3734545}{\sqrt[3]{-964637+36 i \sqrt{136255611}}}-32981\right)}+\frac{1}{2}\surd \left(-\frac{4 \sqrt[3]{-964637+36 i \sqrt{136255611}}}{1083}-\frac{263848}{390963}-\frac{41380}{1083 \sqrt[3]{-964637+36 i \sqrt{136255611}}}+\frac{4000800}{130321 \sqrt{\frac{1}{3} \left(361 \sqrt[3]{-964637+36 i \sqrt{136255611}}+\frac{3734545}{\sqrt[3]{-964637+36 i \sqrt{136255611}}}-32981\right)}}\right.\right.\right.

Guessing that you meant the real root: 1 is the answer ( 4 = 68 17 4=\frac{68}{17} ).

0 Helpful 1 Interesting 0 Brilliant 0 Confused

That is one amusing solution xD

Zhang Xiaokang - 1 year, 11 months ago

The other two roots are the complex roots. Yes, they are complex (pun intended).

A Former Brilliant Member - 1 year, 11 months ago

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