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x² – 26x + 168 = 0
y² – 34y + 285 = 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 'B'. Can you explain this answer?
Most Upvoted Answer
x² – 26x + 168 = 0y² – 34y + 285 = 0a)X > Yb)...
x² – 26x + 168 = 0
x = 12, 14
y² – 34y + 285 = 0
y = 15, 19
x = 12, 14
y² – 34y + 285 = 0
y = 15, 19
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Community Answer
x² – 26x + 168 = 0y² – 34y + 285 = 0a)X > Yb)...
Understanding the Quadratic Equations
To compare the values of X and Y from the given quadratic equations, we first need to solve each equation.
Equation for X
The first equation is:
x² - 26x + 168 = 0
We can use the quadratic formula:
x = [26 ± √(26² - 4 × 1 × 168)] / (2 × 1)
Calculating the discriminant:
- 26² = 676
- 4 × 1 × 168 = 672
- Discriminant = 676 - 672 = 4
Now we can find the roots:
- x = [26 ± √4] / 2
- x = [26 ± 2] / 2
This gives us two roots:
- x₁ = (28 / 2) = 14
- x₂ = (24 / 2) = 12
Thus, X can take the values 12 or 14.
Equation for Y
The second equation is:
y² - 34y + 285 = 0
Using the quadratic formula:
y = [34 ± √(34² - 4 × 1 × 285)] / (2 × 1)
Calculating the discriminant:
- 34² = 1156
- 4 × 1 × 285 = 1140
- Discriminant = 1156 - 1140 = 16
Finding the roots:
- y = [34 ± √16] / 2
- y = [34 ± 4] / 2
This gives us two roots:
- y₁ = (38 / 2) = 19
- y₂ = (30 / 2) = 15
Thus, Y can take the values 15 or 19.
Comparison of X and Y
Now we compare the values of X and Y:
- Maximum value of X = 14
- Minimum value of Y = 15
Since 14 < 15,="" we="" find="" that="" x="" />< />
Conclusion
Therefore, the correct answer is option 'B': X < y.="" />
To compare the values of X and Y from the given quadratic equations, we first need to solve each equation.
Equation for X
The first equation is:
x² - 26x + 168 = 0
We can use the quadratic formula:
x = [26 ± √(26² - 4 × 1 × 168)] / (2 × 1)
Calculating the discriminant:
- 26² = 676
- 4 × 1 × 168 = 672
- Discriminant = 676 - 672 = 4
Now we can find the roots:
- x = [26 ± √4] / 2
- x = [26 ± 2] / 2
This gives us two roots:
- x₁ = (28 / 2) = 14
- x₂ = (24 / 2) = 12
Thus, X can take the values 12 or 14.
Equation for Y
The second equation is:
y² - 34y + 285 = 0
Using the quadratic formula:
y = [34 ± √(34² - 4 × 1 × 285)] / (2 × 1)
Calculating the discriminant:
- 34² = 1156
- 4 × 1 × 285 = 1140
- Discriminant = 1156 - 1140 = 16
Finding the roots:
- y = [34 ± √16] / 2
- y = [34 ± 4] / 2
This gives us two roots:
- y₁ = (38 / 2) = 19
- y₂ = (30 / 2) = 15
Thus, Y can take the values 15 or 19.
Comparison of X and Y
Now we compare the values of X and Y:
- Maximum value of X = 14
- Minimum value of Y = 15
Since 14 < 15,="" we="" find="" that="" x="" />< />
Conclusion
Therefore, the correct answer is option 'B': X < y.="" />
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