Solve the following mathematical equation and find the value of a and b. √a + b = 7 √b + a = 11
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√a + b = 11 => b = 11 - √a
7√b + a = 11
=> 49 b = (11 - a)²
=> 539 - 49 √a = (11 - a)²
=> 49 √a = 539 - (11 - a)²
=> 49² a = 539² + (11 - a)⁴ - 2 * 539 * (11 - a)
=> a⁴ - 4 * 11 a³ + 6 *11² * a² - 4 *11³ a + 11⁴ - 22 * 539 + 1078 a - 49² a = 0
=> P(x) = a⁴ - 44 a³ + 726 a² - 7725 a + 2783 = 0
=>
a has four positive real values at most. It appears.
Two of them are :
a ≈ 0.37304 and a ≈ 27.7359
b ≈ 10.389 and b ≈ 5.7335
There are no real negative roots.
Perhaps there are two more real positive roots or there are two imaginary roots.
The derivative of P(x) = P'(x) = 4 a³ - 132 a² + 1452 a - 7725
has one real root. and two imaginary roots.
So P(x) can have at most two real roots.
The solutions as above are good.
7√b + a = 11
=> 49 b = (11 - a)²
=> 539 - 49 √a = (11 - a)²
=> 49 √a = 539 - (11 - a)²
=> 49² a = 539² + (11 - a)⁴ - 2 * 539 * (11 - a)
=> a⁴ - 4 * 11 a³ + 6 *11² * a² - 4 *11³ a + 11⁴ - 22 * 539 + 1078 a - 49² a = 0
=> P(x) = a⁴ - 44 a³ + 726 a² - 7725 a + 2783 = 0
=>
a has four positive real values at most. It appears.
Two of them are :
a ≈ 0.37304 and a ≈ 27.7359
b ≈ 10.389 and b ≈ 5.7335
There are no real negative roots.
Perhaps there are two more real positive roots or there are two imaginary roots.
The derivative of P(x) = P'(x) = 4 a³ - 132 a² + 1452 a - 7725
has one real root. and two imaginary roots.
So P(x) can have at most two real roots.
The solutions as above are good.
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