Question
I. 5x² - 24 x + 28 = 0   II. 4y² - 8 y -
12= 0Â Â Â In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answer.Solution
I. 5x² - 24 x + 28 = 0    5x² - 14x - 10x + 28 = 0    x (5x – 14) – 2 (5x – 14 )  (x – 2 ) (5x – 14)  x = 14/5, 2  II. 4y² - 8y - 12 = 0   4y²- 12y + 4y - 12 = 0   4 (y – 3) + 4(y – 3 )  (y– 3) (4y + 4)  y = 3, -1  Hence, relationship between x and y cannot be established.  Alternate Method:  If signs of quadratic equation is -ve and +ve respectively then the roots of equation will be +ve and +ve.   So, roots of first equation = x = 14/5, 2  If signs of quadratic equation is -ve and -ve respectively then the roots of equation will be +ve and -ve. (note: -ve sign will come in smaller root)  So, roots of second equation = y = 3, -1  After comparing roots of quadratice eqution we can conclude that relation cannot be established between x and y.
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