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Shruti Patil

· started a discussion

· 1 Months ago

Value of y could be different (Question: 103127.0 )

In this question value of y could be 6 and -1.
And so the value changes the answer to y

Question:

Directions : In the givenquestion two equations numbered I and II are given. You have to solve both the equations and

Give answer if  

I. x2 – 14x + 48 = 0

II. y2 + 6 = 5y

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 – 14x + 48 = 0

⇒ x2 – 8x – 6x + 48 = 0

⇒ x (x – 8) -6 (x – 8) = 0

⇒ (x – 6) (x – 8) = 0

∴ X = 6 or 8

II. y2 -5y + 6 = 0

⇒ y2 – 3y -2y + 6 = 0

⇒ y(y – 3) -2(y – 3) = 0

⇒ (y – 2) (y – 3) = 0

∴ y = 2 or 3

Clearly, x > y

kanhaiya jha

· commented

· 1 Months ago

only one factor you can find for one qudratic....

kanhaiya jha

· commented

· 1 Months ago

No....y=3 and 2

I. x2 – 14x + 48 = 0
⇒ x2 – 8x – 6x + 48 = 0
⇒ x (x – 8) -6 (x – 8) = 0
⇒ (x – 6) (x – 8) = 0
∴ X = 6 or 8
II. y2 -5y + 6 = 0
⇒ y2 – 3y -2y + 6 = 0
⇒ y(y – 3) -2(y – 3) = 0
⇒ (y – 2) (y – 3) = 0
∴ y = 2 or 3
Clearly, x > y

Shruti Patil

· commented

· 1 Months ago

Option B

Shruti Patil

· commented

· 1 Months ago

Option B

Shruti Patil

· commented

· 1 Months ago

Option B
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