Question 125

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

Solution

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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SSC CGL 2014 Tier 1 19 Oct shift 2

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

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