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Solve the question and mark your answer as

  1. a) if x > y b) if x < y
  2. c) if x ≥ y d) if x ≤ y
  3. e) if x = y or relation can’t be established

 

  1. i) x2 – 6x – 27 = 0
  2. ii) y2 – 7y – 60 = 0

Here, no need to solve the quadratic equations Just observe that the constant term in 1st equation is negative [i.e – 27] and also, the constant term in 2nd equation is negative [i.e – 60]. In such cases, we can directly say that relation cannot be established between x and y as one of the roots of 1st equation will be (+)ve and the other one will be (-)ve. Same is the situation for 2nd equation. So, when we compare the values. We get x > y and x < y together.

So, the answer is (e) option.

Given below are the cases that exist in quadratic equations in terms of the signs of the coefficients and we will indicate the signs of the roots of the equation based on signs of coefficients.

 

Here are some of the examples, which you can solve directly just by observing the signs of the coefficients

Solve the questions and mark your answer as

  1. a) if x > y b) if < y
  2. c) if x ≥ y d) if x ≤ y
  3. e) if x = y or relation can’t be established

1) i) x2 – 5x + 6 = 0

  1. ii) y2 + 5y + 6 = 0

Sol:

xy
+
+

So directly, x > y

 

2) i) 21x2 – 13x + 2 = 0

  1. ii) 56y2 + 15y + 1 = 0

Sol:

xy
+
+

Here also, directly x > y

Self evaluation test

1) i) x2 – 11x – 42 = 0

  1. ii) y2 + 15y – 34 = 0

 

2) i) x2 – 12x + 32 = 0

  1. ii) y2 + 9y + 20 = 0

 

3) i) x2 = 36

  1. ii) y2 = 25

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