Instructions

In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and Give answer :

Question 51

I. $$2x^{2} - 17x + 36 = 0$$
II. $$2y^{2} - 19y + 44 = 0$$

Solution

I : $$2x^{2} - 17x + 36 = 0$$

=> $$2x^{2} - 8x - 9x + 36 = 0$$

=> $$2x (x - 4) -9 (x - 4) = 0$$

=> $$(x - 4) (2x - 9) = 0$$

=> $$x = 4, \frac{9}{2}$$

II : $$2y^{2} - 19y + 44 = 0$$

=> $$2y^{2} - 8y - 11y + 44 = 0$$

=> $$2y (y - 4) -11 (y - 4) = 0$$

=> $$(y - 4) (2y - 11) = 0$$

=> $$y = 4, \frac{11}{2}$$

Since, both values of $$y \geq x$$, but $$\frac{9}{2}(x) > 4(y)$$

Hence, no relation can be established.

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I. $$2x^{2} - 17x + 36 = 0$$II. $$2y^{2} - 19y + 44 = 0$$