Question

A²+b²=5; ab=2 знайти: 1) a+b; 2 |a-b|​

Answer 1

Решение задания прилагаю

Answer 2

(a+b)²=a²+2ab+b²=5+4=9 =>

a+b=3

(a-b) ²=a²-2ab+b²=5-4=1 =>

|a-b|=1

Answer 3

Given the equations:

1. $a^2 + b^2 = 5$
2. $ab = 2$

We need to find:

1. $a + b$
2. $|a - b|$

Let's solve these step by step:

First, let's square the equation $a^2 + b^2 = 5$:

$(a^2 + b^2)^2 = 25$

Expanding the left side of the equation:

$a^4 + 2a^2b^2 + b^4 = 25$

Since we know that $ab = 2$, we can substitute this value into the equation:

$a^4 + 2(2^2) + b^4 = 25$

Simplifying:

$a^4 + 4 + b^4 = 25$

Now we have a new equation involving fourth powers of $a$ and $b$. However, we still have the equation $ab = 2$, which can help us find the values of $a$ and $b$. From the second equation, we can express $b$ in terms of $a$:

$b = \frac{2}{a}$

Substitute this value into the equation $a^2 + b^2 = 5$:

$a^2 + \left(\frac{2}{a}\right)^2 = 5$

Simplify:

$a^4 - 5a^2 + 4 = 0$

This is a quadratic equation in terms of $a^2$. Factoring it:

$(a^2 - 4)(a^2 - 1) = 0$

This gives us two sets of possible solutions:

1. $a^2 - 4 = 0$: $a = \pm 2$
2. $a^2 - 1 = 0$: $a = \pm 1$

For each value of $a$, we can calculate the corresponding value of $b$ using $b = \frac{2}{a}$.

When $a = 2$, $b = 1$. When $a = -2$, $b = -1$. When $a = 1$, $b = 2$. When $a = -1$, $b = -2$.

Now, we can calculate the requested values:

1. $a + b$ for all cases:

   • When $a = 2$, $b = 1$: $a + b = 2 + 1 = 3$
   • When $a = -2$, $b = -1$: $a + b = -2 + (-1) = -3$
   • When $a = 1$, $b = 2$: $a + b = 1 + 2 = 3$
   • When $a = -1$, $b = -2$: $a + b = -1 + (-2) = -3$

2. $|a - b|$ for all cases:

   • When $a = 2$, $b = 1$: $|a - b| = |2 - 1| = 1$
   • When $a = -2$, $b = -1$: $|a - b| = |-2 - (-1)| = |-1| =$