if x = 1/(2-√3), show that the value of (x³ - 2x² - 7x + 5) is 3
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x=1/(2-rt3) = 2+rt3/(2-rt3)*(2+rt3) = 2+rt3 [rationalizing]
x^3 = (2+rt3)^3 = 8+3rt3+6rt3(2+rt3) = 8 + 3rt3 +12rt3 + 18= 8+15rt3+18
x^2= (2+rt3)^2 = 4 +3 + 4rt3 = 7 + 4rt3
Now, x^3 - 2x^2 - 7x + 5 = 8 + 15rt3 + 18 - 2*(7 + 4rt3) - 7*(2+rt3) + 5
= 8 + 15rt3 + 18 - 14 - 8rt3 - 14 - 7rt3 + 5 = 3...Hence proved
N.B [a+b]^3 = a^3 + b^3 + 3ab(a+b)
x^3 = (2+rt3)^3 = 8+3rt3+6rt3(2+rt3) = 8 + 3rt3 +12rt3 + 18= 8+15rt3+18
x^2= (2+rt3)^2 = 4 +3 + 4rt3 = 7 + 4rt3
Now, x^3 - 2x^2 - 7x + 5 = 8 + 15rt3 + 18 - 2*(7 + 4rt3) - 7*(2+rt3) + 5
= 8 + 15rt3 + 18 - 14 - 8rt3 - 14 - 7rt3 + 5 = 3...Hence proved
N.B [a+b]^3 = a^3 + b^3 + 3ab(a+b)
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