Question

If x+y=12 and xy =14 find x 2 + y 2

Answer 1

Answer:

The value of $x^2+y^2$ is equal to 116.

To find:

The value of $x^2+y^2$

Solution:

Given that the value of the sum of x and y is 12 and the product of the terms x and y is 14.

$x+y=12\\\\xy=14$

$x^2+y^2=?$

We know that the value of  

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

Using  this identity $(x+y)^2$ becomes,

$(x+y)^2=x^2+y^2+2xy$

Substitute x + y = 12 and xy = 14 in the above expansion, we get,

$\begin{array} { l } { ( 12 ) ^ { 2 } = x ^ { 2 } + y ^ { 2 } + 2 \times 14 } \\\\ { 144 = x ^ { 2 } + y ^ { 2 } + 28 } \\\\ { x ^ { 2 } + y ^ { 2 } = 144 - 28 } \\\\ { x ^ { 2 } + y ^ { 2 } = 116 } \end{array}$

Therefore, the value of $x^2+y^2$ is equal to 116.

Answer 2

Answer:

x²+y² = 116

Step-by-step explanation:

Given x+y=12 ---(1)

xy = 14 ------(2)

On Squaring both sides of equation (1) ,we get

(x+y)²= 12²

=> x²+2xy+y² = 144

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By algebraic identity:

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

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=> x²+y² +2×14=144

/* From (2)*/

=> x²+y²+28 = 144

=> x²+y² = 144-28

=> x²+y² = 116

Therefore

x²+y² = 116

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