1. If x = √5+ 2, then the value $$\frac{2x^2-3x-2}{3x^2-4x-3}$$ is equal to





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  • By: anil on 05 May 2019 02.57 pm
    Given , x = (√5+ 2)
    $$frac{2x^2-3x-2}{3x^2-4x-3}$$ = $$frac{2(sqrt{5}+ 2)^2-3(sqrt{5}+ 2)-2}{3(sqrt{5}+ 2)^2-4(sqrt{5}+ 2)-3}$$
    = $$frac{21.18}{33.88}$$
    = 0.625
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