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x² – 30x + 221 = 0
y² – 33y + 270 = 0
  • a)
    X > Y
  • b)
    X < Y
  • c)
    X ≥ Y
  • d)
    X ≤ Y
  • e)
    X = Y or relation cannot be established
Correct answer is option 'E'. Can you explain this answer?
Most Upvoted Answer
x² – 30x + 221 = 0y² – 33y + 270 = 0a)X > Yb)...
x² – 30x + 221 = 0
x = 13, 17
y² – 33y + 270 = 0
y = 15, 18
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Community Answer
x² – 30x + 221 = 0y² – 33y + 270 = 0a)X > Yb)...
Understanding the Quadratic Equations
The problem involves solving two quadratic equations to compare their roots, X and Y.
Equations Given:
- X: \( x^2 - 30x + 221 = 0 \)
- Y: \( y^2 - 33y + 270 = 0 \)
Finding Roots of X:
To find the roots of the equation for X, we use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
For X:
- \( a = 1 \)
- \( b = -30 \)
- \( c = 221 \)
Calculating the discriminant:
\[
D_x = (-30)^2 - 4 \cdot 1 \cdot 221 = 900 - 884 = 16
\]
Thus, the roots for X are:
\[
x = \frac{30 \pm 4}{2} \implies x = 17 \text{ or } x = 13
\]
Finding Roots of Y:
Using the same formula for Y:
For Y:
- \( a = 1 \)
- \( b = -33 \)
- \( c = 270 \)
Calculating the discriminant:
\[
D_y = (-33)^2 - 4 \cdot 1 \cdot 270 = 1089 - 1080 = 9
\]
Thus, the roots for Y are:
\[
y = \frac{33 \pm 3}{2} \implies y = 18 \text{ or } y = 15
\]
Comparing the Roots:
Now, we have:
- Roots of X: 17, 13
- Roots of Y: 18, 15
When comparing:
- Maximum X (17) < maximum="" y="" />
- Minimum X (13) < minimum="" y="" />
Conclusion:
Since both roots of X are less than the roots of Y, we can conclude that there is no consistent relationship established between X and Y that would lead to a definitive inequality or equality, hence:
Correct Answer: e) X = Y or relation cannot be established.
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x² – 30x + 221 = 0y² – 33y + 270 = 0a)X > Yb)X < Yc)X ≥ Yd)X ≤ Ye)X = Y or relation cannot be establishedCorrect answer is option 'E'. Can you explain this answer?
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