Question

A square+ B Square and (A+B) Square when A- B=2 and AB = 63

Answer 1

Answer:

Step-by-step explanation:

Question:  Find

(i) $a^{2} + b^{2}$

(ii) $(a+b)^{2}$

where $a - b = 2$ and $ab = 63$

(i)

$(a-b)^{2}$ is the identity to be used.

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

Before proceeding any further, we can see that $a^{2} + b^{2}$ is part of this identity.  Now, we have to substitute the numerical values in place of the variables in the identity. (Substitute numbers instead of a and b.)

Since $a-b = 2$, we can substitute it in place of $a-b$ in the identity.  

$(2)^{2}$ = a^{2} + b^{2} - 2ab[/tex]

Next, we don't know $a^{2} + b^{2}$.  It can be left as it is.

Now, we know that $ab$ is$63$.  Therefore, $2ab$ =$126$.  ($63$×$2$).

The equation now reads:

$(2)^{2} = a^{2} +b^{2}  -126$.

This can be further simplified as:

$4 = a^{2} +b^{2}  -126$.

Now, add 126 to RHS and add it to  the LHS (to cancel 126 out from RHS...variables on one side, constants on the other.)

You are left with:

$130 = a^{2} +b^{2}$.

(ii)

Here, we already know all the terms, (except $(a+b)^{2}$) so we can substitute them in the following identity.

ID:  $(a+b)^{2}$ = a^{2} + b^{2} + 2ab[/tex]

$(a+b)^{2}  = 130 + 126\\</p><p>(a+b)^{2} = 256$

(To find a + b)

We know that:

$(a+b)^{2}  = 130 + 126\\ </p><p>All we have to do is find the square root of [tex](a+b)^{2}$, which is$a+b$. ($(a+b)$×$(a+b)$)

Similarly, we have to find the square root of [tex]256, which is 16.

Therefore,

a + b = 16.