Instructions

In these questions, two equations numbered I and II are given. You have to solve both the equations and mark the appropriate answer. Give answer

Question 57

I. $$16x^{2} - 14x + 3 = 0$$
II. $$6y^{2} - 19y + 15 = 0$$

Solution

I : $$16x^{2} - 14x + 3 = 0$$

=> $$16x^{2} - 8x - 6x + 3 = 0$$

=> $$8x (2x - 1) -3 (2x - 1) = 0$$

=> $$(8x - 3) (2x - 1) = 0$$

=> $$x = \frac{3}{8}, \frac{1}{2}$$

II : $$6y^{2} - 19y + 15 = 0$$

=> $$6y^{2} - 9y - 10y + 15 = 0$$

=> $$3y (2y - 3) -5 (2y - 3) = 0$$

=> $$(3y - 5) (2y - 3) = 0$$

=> $$y = \frac{5}{3}, \frac{3}{2}$$

Since, both values of x < 1 and both values of y > 1.

=> y > x

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I. $$16x^{2} - 14x + 3 = 0$$II. $$6y^{2} - 19y + 15 = 0$$