Don't use a calculating machine

Algebra Level pending

k 6 − 11 k 3 = 80 \large k^{ 6 }-11k^{ 3 }=80

Find the positive value to two decimal places of k k in the equation above.


The answer is 2.52.

Nikolas Кraj by Nikolas Кraj , 16, Albania

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1 solution

Sathvik Acharya
Dec 19, 2020

k 6 − 11 k 3 − 80 = 0 k^6-11k^3-80=0 Substituting x = k 3 x=k^3 in the above equation, x 2 − 11 x − 80 = 0 x^2-11x-80=0 ⟹ ( x − 16 ) ( x + 5 ) = 0 \implies (x-16)(x+5)=0 ⟹ x = 16 or x = − 5 \implies x=16\; \; \text{or}\; \; x=-5 Since k > 0 , x = 16 k>0, x=16 , ⟹ k = x 3 = 16 3 ≈ 2.52 \implies k=\sqrt[3]{x}=\boxed{\sqrt[3]{16}\approx 2.52}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

it's a nice task and can be done without calculator. besides the 16 3 \sqrt[3]{16} part.

num IC - 5 months, 1 week ago

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