Question

Q.1. Solve p = (z+py)^2​

Answer 1

Step-by-step explanation:

p= ( z + py)²

p= z²+p²y²+ 2zpy

p-2zpy= z²+p²y²

p(1-2zy)= z²+p²y²

p = z²+p²y² / 1-2zy

Answer 2

Answer:

The algebraic expression for the given equation is $p = z^{2} + p^{2} y^{2} +2zpy$

Step-by-step explanation:

Given: $p = (z+py)^{2}$

To find: Simplify the given equation $p = (z+py)^{2}$

Formula used: $(a+b)^{2} = a^{2} +b^{2} +2ab$

Solution:

The given equation is in the form of a and b

Given equation is $p = (z+py)^{2}$

Consider,

a = z;

b = py;

As we know that,

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

The above formula is used to factorize the method in simple terms.

It is a binomial term and also the major algebraic identities.

Therefore, now replace the value in the formula.

$(z+py)^{2} = z^{2} + (py)^{2} +2(z)(py)$

$(z+py)^{2} = z^{2} + p^{2} y^{2} +2zpy$

After simplify the algebraic expression we get,

$p = z^{2} + p^{2} y^{2} +2zpy$

Final answer:

The final  algebraic expression for the given equation is $p = z^{2} + p^{2} y^{2} +2zpy$

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