# Solve the following mathematical equation and find the value of a and b. √a + b = 7 √b + a = 11

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by poorniaaught

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by poorniaaught

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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 :

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.