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rishimangal
Joined: 20 Jul 2025
Last visit: 28 Apr 2026
Posts: 3
Given Kudos: 1
Location: India
Schools: LBS '26 (A) INSEAD Aug '26 (WA)
GMAT Focus 1: 675 Q85 V86 DI79 
GPA: 3.1
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23 Jul 2025, 12:09
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I like the solution - it’s helpful.
27 Jul 2025, 05:33
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How I solved it was x^4 - x^3 - 6x^2 = 0 now as per the normal formula sum of roots is -b/a which is 1 so I marked that answer - why is that wrong
Bunuel
27 Jul 2025, 09:12
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Pikachu007
First of all, we have a 4th-degree equation, not a quadratic.
We get: \(x^2(x^2 - x - 6) = 0\). So, \(x = 0\), or solve \(x^2 - x - 6 = 0\), which gives \(x = 3\) and \(x = -2\).
Yes, the sum of the roots of \(x^2 - x - 6\) is 1, but as shown in the solution, \(x = -2\) is not valid because it does not satisfy the original equation \(x = \sqrt[4]{x^3 + 6x^2}\). The right-hand side is always non-negative, so \(x\) must be \(\geq 0\).
Please review the discussion for more clarity.
