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    Question

    I. 5x² - 24 x + 28 = 0     II. 4y² - 8 y  -

    12= 0     In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answer.
    A If x ≥ y Correct Answer Incorrect Answer
    B If x ≤ y Correct Answer Incorrect Answer
    C If x > y Correct Answer Incorrect Answer
    D If x=y or relationship between x and y cannot be established Correct Answer Incorrect 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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