Question

ans this question pls​

Answer 1

Answer:

2or1

Step-by-step explanation:

this is the answer

Answer 2

1 and 2

Step-by-step explanation:

Given:

$x + \sqrt{5 - {x}^{2} } = 3$

To find:

The value of x

Solution:

Point to be noted:

1st There are square root in the question so we will square the terms as required.

2nd After elimination of roots we will follow the equation rule i.e writing the veriable in one side and constant in one side and just Simplify.

$x + \sqrt{5 - {x}^{2} } = 3 \\ \\ \sqrt{5 - {x}^{2} } = 3 - x$

[Squaring both side]

${(\sqrt{5 - {x}^{2} } )}^{2} ={( 3 - x) } {}^{2} \\ \\5 - {x}^{2} = {(3)}^{2} - 2.3.x + {x}^{2} \\ \\ 5 - {x}^{2} = 9 - 6x + {x}^{2} \\ \\ 0 = {x}^{2} + {x}^{2} - 6x + 9 - 5 \\ \\ 2 {x}^{2} - 6x + 4 = 0$

[Factorising the equation]

$2 {x}^{2} - 6x + 4 = 0 \\ \\ 2 {x}^{2} - 2x - 4x + 4 = 0 \\ \\ 2x(x - 1) - 4(x - 1) = 0\\ \\ (x - 1)(2x - 4) = 0 \\ \\ \text{either,} \: \: \: \: x - 1 = 0 \implies x = 1 \\ \\ \text{or,} \: \: \: 2x - 4 = 0 \\ \\ 2x = 4 \implies \: x \: = 2$

Therefore, the required answer is 1 and 2.

Some Important Algebra Formula

${(x + y)}^{2}={x}^{2}+{y}^{2}+2xy\\ \\{(x - y)}^{2}={x}^{2}+{y}^{2}-2xy\\ \\ (x + y)^{2}+(x-y)^{2}=2( {x}^{2}+{y}^{2} )\\ \\(x+y)^{2}-(x-y) {}^{2}=4xy$