Question

Given that a + b = 9 and

a^2 + b^2 =41 find the value of a − b using algebraic method.

Answer 1

Answer:

a=5

b=4

Step-by-step explanation:

Given,

a+b=9

a²+b²= 41

a - b = (x)

Now, We can radomly use every number to get the sum 9 and try to get the answer

5 + 4 = 9

5² + 4² = 41

Hope it proves

at last,

a-b

5 - 4 = 1

Hence Proved

Answer 2

$\large \underline{ \underline{ \text{Question:}}}$

• If a + b = 9 and a² + b² = 41. Then find the value of a - b using algebraic method.

$\large \underline{ \underline{ \text{Solution:}}}$

Given that,

• $a + b = 9 \: \: \: \: \: ...(1)$

And,

• ${a}^{2} + {b}^{2} = 41 \: \: \: \: \: ...(2)$

Let take Equation [1],

$\implies a + b = 9 \: \: \: \: \: ...(1)$

Squaring both sides,

$\implies {(a + b)}^{2} = {(9)}^{2}$

According to Identity 1;

$[ {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy ]$

Let us solve,

$\implies {a}^{2} + {b}^{2} + 2ab = 81$

By Equation [2],

$\implies 41 + 2ab = 81$

Finding the value of 2ab,

$\implies 2ab = 81 - 41$

$\implies 2ab =40 \: \: \: \: \: ...(3)$

According to Identity 2;

$[ {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy ]$

Let us find the value of a - b,

$\implies {(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab$

By Equation [2] and [3],

$\implies {(a - b)}^{2} = 41 - 40$

$\implies {(a - b)}^{2} = 1$

$\implies {(a - b)}^{2} = {1}^{2}$

Comparing both sides,

$\implies a - b = 1$

Therefore,

• The value of a - b is 1.

$\\ \large \underline{ \underline{ \text{Required Answer:}}}$

• The value of a - b is 1.