In each question two equations numbered I and II are given. You have to solve both the equations and mark the answer. I. 4x2 – (8 + √10)x + 2√10 = 0 II. 2y2 – (4 + 3√11)y + 6√11 = 0
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In each question two equations numbered I and II are given. You have to solve both the equations and mark the answer. I. 4x2 – (8 + √10)x + 2√10 = 0 II. 2y2 – (4 + 3√11)y + 6√11 = 0
if x > y
if x ≥ y
if x < y
if x ≤ y
if x = y or the relation between x and y can't be determined
Equation I: 4x^2 - 8x - sqrt(10)x + 2sqrt(10) = 4x(x-2) - sqrt(10)(x-2) = (4x-sqrt(10))(x-2) = 0. Roots: x = 2, x = sqrt(10)/4 approx 0.79. Equation II: 2y^2 - 4y - 3sqrt(11)y + 6sqrt(11) = 2y(y-2) - 3sqrt(11)(y-2) = (2y-3sqrt(11))(y-2) = 0. Roots: y = 2, y = 3sqrt(11)/2 approx 4.97. Comparing roots, x <= y.
Factoring equation I gives (2x minus the square root of 10)(2x minus 4) equals 0, so the roots for x are the square root of 10 over 2 and 2. Factoring equation II gives (2y minus the square root of 11)(y minus 3 times the square root of 11) equals 0, resulting in y values of the square root of 11 over 2 and 3 times the square root of 11. Since the square root of 10 over 2 is less than both y roots, and the value 2 is less than the square root of 11 over 2, every root of x is smaller than every root of y. Therefore, x is less than or equal to y.