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Tarun

· started a discussion

· 1 Months ago

One more solution (Question: 99897.3 )

The roots y can also be -1 and -(15/2), if we take dividing roots as -2 and 15.

Question:
DIRECTIONS: In the given question two equations numbered I and II are given. You have to solve both the equations.

I. x2 – 6x = 7


II. 2y2 + 13y + 15 = 0

Options:
A) x > y
B) x ≥ y
C) x < y
D) x ≤ y
E) x = y or the relationship cannot be established
Solution:

I. x2 – 6x = 7

⇒ x2 – 7x + x – 7 = 0

⇒ x(x – 7) + 1(x – 7) = 0

⇒ (x + 1) (x – 7) = 0

⇒ x = -1 or 7

II. 2y2 + 13y + 15 = 0

⇒ 2y2 + 10y + 3y + 15 = 0

⇒ 2y (y + 5) + 3(y + 5) = 0

⇒ (y + 5) (2y + 3) = 0

⇒ y = -5 or - \(\cfrac{3}{2}\)

Knowledge Expert

· commented

· 1 Months ago

Did you see ever that a quadratic equation has four solution??????

We didn't.................Please check your roots by putting in the equation whether it satisfies or not.....

Your solution doesn't satisfy the equation....so your solution is totally wrong......
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