Question 136

 I. $$2x^{2}+18x+40=0$$
II. $$2y^{2} +15y+27=0$$

Solution

I.$$2x^{2} + 18x + 40 = 0$$

=> $$2x^2 + 8x + 10x + 40 = 0$$

=> $$2x (x + 4) + 10 (x + 4) = 0$$

=> $$(x + 4) (2x + 10) = 0$$

=> $$x = -4 , -5$$

II.$$2y^{2} + 15y + 27 = 0$$

=> $$2y^2 + 6y + 9y + 27 = 0$$

=> $$2y (y + 3) + 9 (y + 3) = 0$$

=> $$(y + 3) (2y + 9) = 0$$

=> $$y = -3 , \frac{-9}{2}$$

$$\therefore$$ No relation can be established.


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