Question

If a^2 + b^2 = (a+b)^2 then why is it = (a+b)^2 -2ab in this answer for this question:


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Answer 1

Step-by-step explanation:

Actually a²+b²+2ab=(a+b)²

So, a²+b²=(a+b)²-2ab

Answer 2

Step-by-step explanation:

⠀⠀⠀⠀★ a² + b² = (a + b)²

A big "NO"! (a² + b²) is not equivalent to (a + b)². Let's understand this how! Firstly, we need to prove a² + b² ≠ (a + b)² .And, then we'll prove how a² + b² = (a + b)² – 2ab.

$\twoheadrightarrow \sf{\quad { (a + b)^2}}$

Basically, (a + b)² is the square of (a + b). That is has been multiplied with itself. Thus, we can also write it as,

$\twoheadrightarrow \sf{\quad { (a + b)(a + b)}}$

We have studied that, to multiply two binomials, we use distributive law of multiplication over addition two times. That means, each term of the first binomial with each term with the second binomial.

$\twoheadrightarrow \sf{\quad { a(a + b) + b(a + b)}}$

Once again , using distributive law of multiplication over addition.

$\twoheadrightarrow \sf{\quad { a(a) + a(b) + b(a) + b(b)}}$

Now, performing multiplication as usual. a(a) and b(b) can be written as a² and b².

$\twoheadrightarrow \sf{\quad { a^2 + ab + ab + b^2}}$

Adding the like terms.

$\twoheadrightarrow \quad\boxed{\sf { a^2 + 2ab + b^2}}$

As we get that, (a + b)² = a² + b² + 2ab. Thus, it has been proven that a² + b² ≠ (a + b)².

Let' understand with an example too. Suppose we have the value of a as 2 and value of b as 3.

»⠀⠀a² + b² ≠ (a + b)²

Now, in the LHS :

↠⠀⠀a² + b²

↠⠀⠀(2)² + (3)²

↠⠀⠀4 + 9

↠⠀⠀13

Whereas, in the RHS :

↠⠀⠀(a + b)²

↠⠀⠀(2 + 3)²

↠⠀⠀(5)²

↠⠀⠀25

So, 13 ≠ 25 ; LHS ≠ RHS and a² + b² ≠ (a + b)².

⠀⠀________________________________⠀⠀⠀⠀

Now, we have to prove that how,

⠀⠀⠀⠀★ a² + b² = (a + b)² – 2ab

Basically, in the above conclusion we got that (a + b)² = a² + b² + 2ab. In the LHS, we have a² + b². So, substitute the value of (a + b)² in the RHS.

$\twoheadrightarrow \sf{\quad { (a + b)^2 - 2ab}}$

Substitute the value of (a + b)².

$\twoheadrightarrow \sf{\quad { a^2 + b^2 \cancel{+ 2ab - 2ab}}}$

+2ab and – 2ab will get cancelled.

$\twoheadrightarrow \sf{\quad { a^2 + b^2 \dots \; R.H.S }}$

Thus, we get that,

↠⠀⠀LHS = RHS

Hence, proved!

$\twoheadrightarrow \quad\boxed{\sf { a^2+ b^2 =(a + b)^2 - 2ab }}$