The sum of two irrational numbers multiplied by a larger one is 70 and their difference is multiplied by the smaller one is 12 the two numbers are
Answers
Answer:
5√2 and 2√2
Step-by-step explanation:
Let the two irrational numbers be √a and √b, where a and b are positive integers but not a perfect square with √a > √b
By the given conditions,
√a (√a + √b) = 70
or, a + √(ab) = 70 ..... (1)
√b (√a - √b) = 12
or, √(ab) - b = 12 .... (2)
Now {(1) - (2)} gives
a + b = 58
or, a = 58 - b
Putting a = 58 - b in (1), we get
58 - b + √{b (58 - b)} = 70
or, 58 - b + √(58b - b²) = 70
or, √(58b - b²) = b + 12
or, 58b - b² = b² + 24b + 144
or, 2b² - 34b + 144 = 0
or, b² - 17b + 72 = 0
or, b² - 9b - 8b + 72 = 0
or, b (b - 9) - 8 (b - 9) = 0
or, (b - 9) (b - 8) = 0
Either b - 9 = 0 or, b - 8 = 0
This gives b = 9, 8
Since b is not a perfect square, we take b = 8
Then a = 58 - 8 = 50
So √a = √50 = 5√2
and √b = √8 = 2√2
Therefore the two irrational numbers are 5√2 and 2√2
Answer:
The numbers are 5 √(2), 2 √(2)
Step-by-step explanation:
Given -
The sum of two irrational numbers multiplied by a larger one is 70.
Their difference is multiplied by the smaller one is 12.
To Find -
The two numbers.
Solution -
Consider the -
The numbers as - x and y.
∴ x > y
We have,
⇒ (x + y)(x) = 70 or x² + xy = 70 ......Eq (1)
⇒ (x - y)(y) = 12 or xy - y² = 12 ......Eq (2)
Adding Eq 1 & 2 -
⇒ x² + y² = 58 .....Eq (3)
⇒ y =√(58 - x²)
Now,
Subsituting the value of above expression in Eq 1 -
⇒ x² + x √(58 - x²) = 70
⇒ x √(58 - x²) = 70 - x²
Taking Squares on both sides,
⇒ x² (58 - x²) = 4900 - 140 x² + x⁴
⇒ 58 x² - x⁴ = 4900 - 140 x² + x⁴
⇒ 4900 - 140 x² + x⁴ = 58 x² - x⁴
⇒ 2x⁴ - 198x² + 4900 = 0
⇒ x⁴ - 99x² + 2450 = 0
⇒ x² = [99 ± √(99² - 4 × 1 × 2450)]/2
⇒ x² = (99 ± 1)/2
⇒ x² = 50, 49
⇒ x = 5 √(2)
⇒ x = 7
We aren't considering it 7.
∵ Because it was given in question that they are irrational numbers.
From Eq 3 -
⇒ y² = 58 - x² = 58 - 50 = 8
⇒ y = 2 √(2)
∴ The numbers are 5 √(2), 2 √(2)