Question

root a + root b ka whole square ​

Answer 1

Answer:

$(\sqrt{a} + \sqrt{b} )^2$ = $a+ 2\sqrt{ab} +b$ is the required answer.

Step-by-step explanation:

Explanation:

Given that, $(\sqrt{a} + \sqrt{b} )^2$

According to the question we need to find out the value of, the whole square of  (root a + root b)

• Now, as we know that the property, The algebraic identity used to calculate the square of a pair of numbers is $(a +b)^2$.

• Simply multiplying (a + b) (a + b) will yield the formula for the binomial of the form $(a +b)^2$

• And also formula of $(a+b)^2$ = $a^2 + 2ab+b^2$

Step 1:

Now, from the formula we have,

a = $\sqrt{a}$ , b = $\sqrt{b}$

⇒ $(\sqrt{a} + \sqrt{b} )^2$ =  $(\sqrt{a} )^2 + 2\sqrt{a} \sqrt{b} + (\sqrt{b} )^2$

⇒ $a+ 2\sqrt{ab} +b$

Therefore, $(\sqrt{a} + \sqrt{b} )^2$ = $a+ 2\sqrt{ab} +b$

Final answer:

Hence, $a+ 2\sqrt{ab} +b$ is the required value of $(\sqrt{a} + \sqrt{b} )^2$.

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Answer 2

Answer:

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

Step-by-step explanation:

Given:

we are given that

$(\sqrt a+\sqrt b)^2$

To Find :

Simplify the given expression using suitable identity

Solution :

given expression

$(\sqrt a+\sqrt b)^2$

since we know that $(x+y)^2=x^2+y^2+2xy$

applying the following identity we get

$(\sqrt a+\sqrt b)^2=(\sqrt a)^2+(\sqrt b)^2+2(\sqrt a)(\sqrt b)$

                  $= a+b+2\sqrt{ ab}$

Hence we have simplified that the value of

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

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