Learn and know the formula of ** a plus b whole square** in algebra. This is the first formula that we are going to learn in algebra chapter.

If we learn this formula, we can easily get the formula for a minus b whole square. Still one more formula also we can get i.e. *a* square plus* b* square ( { a }^{ 2 } + { b }^{ 2 } ). So now we will learn what is ** a plus b whole square formula** along with that we will also learn the proof for this formula.

*a* plus *b* whole square formula as follows:

*a*plus

*b*whole square formula as follows:

** a plus b whole square** mathematically written as { \left( a+b \right) }^{ 2 } .

*a*plus

*b*whole square { \left( a+b \right) }^{ 2 } is equal to the

*a*square ( { a }^{ 2 } ) plus the product of 2,

*a*and

*b*(2

*ab*) plus

*b*square ( { b }^{ 2 } ).

Therefore, the formula of { \left( a+b \right) }^{ 2 } = { a }^{ 2 } + 2 × a × b + { b }^{ 2 } . This is one of the identities in algebra.

**Derivation (proof) of ***a* plus *b* whole square formula:

*a*plus

*b*whole square formula:

Along with knowing the formula of ** a plus b whole square, **we should also know how we got that formula i.e. proof.

We know that we can write,

{ \left( a+b \right) }^{ 2 } = (a+b) × (a+b)

= *a *× *a* + *a* × *b* + *b* ×* a* + *b* × *b* (use multiplication of binomial with another binomial concept)

= { a }^{ 2 } + 2ab + { b }^{ 2 }

Therefore, { \left( a+b \right) }^{ 2 } = { a }^{ 2 } + 2 × a × b + { b }^{ 2 }

Form the above formula, we can get { a }^{ 2 } + { b }^{ 2 } = { \left( a+b \right) }^{ 2 } – 2 × *a* × *b*.

Note:

On this formula we can do one mathematics project also.

**Examples:**

Solve { \left( 3x+5y \right) }^{ 2 } ^{
}

Solution:

We know that the given problem is looking like ** a plus b whole square** formula.

So { \left( a+b \right) }^{ 2 } = { a }^{ 2 } + 2 × a × b + { b }^{ 2 }

In this problem, we know that* a* = 3*x* and b = 5*y*

Therefore

{ \left( 3x+5y \right) }^{ 2 }

= { (3x) }^{ 2 } + 2 × 3*x* × 5*y* + { (5y) }^{ 2 } ^{
}

= 9 { x }^{ 2 } + 30*xy* + 25 { y }^{ 2 }

Therefore, { \left( 3x+5y \right) }^{ 2 } = 9 { x }^{ 2 } + 30*xy* + 25 { y }^{ 2 }

Solve: { \left( 4a+3 \right) }^{ 2 }

Solution:

According to the { \left( a+b \right) }^{ 2 } formula,

{ \left( 4a+3 \right) }^{ 2 }

= { (4a) }^{ 2 } + 2 × 4*a* × 3 + { 3 }^{ 2 }

= 16 { a }^{ 2 } + 24*a* + 9

Therefore, { \left( 4a+3 \right) }^{ 2 } = 16 { a }^{ 2 } + 24*a* + 9.

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