Instructions

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

Question 50

I. $$3x^{2}-13x+12=0$$
II. $$2y^{2}-15y+28=0$$

Solution

$$3x^2-13x+12 = 0$$
$$(3x-4)(x-3) = 0$$
$$x = \frac{4}{3}, 3$$

$$2y^2-15y+28 = 0$$
$$(2y-7)(y-4) = 0$$
$$y = \frac{7}{2}, 4$$

x < y

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I. $$3x^{2}-13x+12=0$$II. $$2y^{2}-15y+28=0$$