Question 135

 I.    $$ 2x^{2}+15x+28=0 $$
II.    $$4y^{2} +18y+14=0 $$

Solution

I.$$2x^{2} + 15x + 28 = 0$$

=> $$2x^2 + 8x + 7x + 28 = 0$$

=> $$2x (x + 4) + 7 (x + 4) = 0$$

=> $$(x + 4) (2x + 7) = 0$$

=> $$x = -4 , \frac{-7}{2}$$

II.$$4y^{2} + 18y + 14 = 0$$

=> $$4y^2 + 4y + 14y + 14 = 0$$

=> $$4y (y + 1) + 14 (y +1) = 0$$

=> $$(y + 1) (4y + 14) = 0$$

=> $$y = -1 , \frac{-7}{2}$$

$$\therefore x \leq y$$


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