please solve this sum and give me the solution
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We have x² + 1/(4x²) = 8.
But x² + 1/(4x²) = (x + 1/2x)² - 2*x*1/2x => (x + 1/2x)² -1 = 8
=> (x + 1/2x)² = 9
=> x + 1/2x = ±3
In both cases obtained value of x is real.
Now let S = x³ + 1/8x³ = (x + 1/2x)³ - 3*(x)*(1/2x)*(x+1/2x) = (x + 1/2x)³ - (3/2)*(x+1/2x)
S = (+3)³ - (3/2)(+3) or S = (-3)³ - (3/2)(-3)
=> S = 27 - 9/2 or -27+ 9/2
=> S = ± 22.5
So x³ + 1/8x³ = ± 22.5
khushi2908:
thanks bhaiya
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