The question is about simplifying algebraic expressions involving square roots, specifically exercises…

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The question is about simplifying algebraic expressions involving square roots, specifically exercises labeled '11' and '12' which include calculating xy, (x-y), (x+y) for given values of x and y, and proving algebraic identities like (a+b)2 - (a-b)2 = 4ab.

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Here are the solutions to the exercises.

:
Given $y = \sqrt{2\sqrt{5}-19}$ and $x = \sqrt{2\sqrt{5}+19}$ .
Note that $2\sqrt{5} \approx 4.47$ . Since $2\sqrt{5}-19$ is negative, $y$ is a complex number. We use $i = \sqrt{-1}$ .
So, $y = \sqrt{-(19-2\sqrt{5})} = i\sqrt{19-2\sqrt{5}}$ .
And $x = \sqrt{19+2\sqrt{5}}$ .

a) Calculate $xy$

Substitute the expressions for $x$ and $y$ .
$xy = \sqrt{19+2\sqrt{5}} \cdot \sqrt{2\sqrt{5}-19}$
Rewrite $\sqrt{2\sqrt{5}-19}$ using $i$ .
$\sqrt{2\sqrt{5}-19} = \sqrt{-(19-2\sqrt{5})} = i\sqrt{19-2\sqrt{5}}$
Substitute and multiply.
$xy = \sqrt{19+2\sqrt{5}} \cdot i\sqrt{19-2\sqrt{5}}$ $xy = i \sqrt{(19+2\sqrt{5})(19-2\sqrt{5})}$
Apply the difference of squares formula $(a+b)(a-b) = a^2-b^2$ .
$xy = i \sqrt{19^2 - (2\sqrt{5})^2}$ $xy = i \sqrt{361 - (4 \cdot 5)}$ $xy = i \sqrt{361 - 20}$ $xy = i \sqrt{341}$
The value of $xy$ is $\boxed{i\sqrt{341}}$ .

b) Calculate $(x-y)^2$
Expand $(x-y)^2$ .
$(x-y)^2 = x^2 - 2xy + y^2$
Substitute $x^2$ , $y^2$ , and $xy$ .
$x^2 = (\sqrt{19+2\sqrt{5}})^2 = 19+2\sqrt{5}$
$y^2 = (\sqrt{2\sqrt{5}-19})^2 = 2\sqrt{5}-19$
$xy = i\sqrt{341}$
$(x-y)^2 = (19+2\sqrt{5}) - 2(i\sqrt{341}) + (2\sqrt{5}-19)$
Simplify the expression.
$(x-y)^2 = 19+2\sqrt{5} - 2i\sqrt{341} + 2\sqrt{5}-19$ $(x-y)^2 = (19-19) + (2\sqrt{5}+2\sqrt{5}) - 2i\sqrt{341}$ $(x-y)^2 = 4\sqrt{5} - 2i\sqrt{341}$
The value of $(x-y)^2$ is $\boxed{4\sqrt{5} - 2i\sqrt{341}}$ .

c) Calculate $(x+y)^2$
Expand $(x+y)^2$ .
$(x+y)^2 = x^2 + 2xy + y^2$
Substitute $x^2$ , $y^2$ , and $xy$ .
$(x+y)^2 = (19+2\sqrt{5}) + 2(i\sqrt{341}) + (2\sqrt{5}-19)$
Simplify the expression.
$(x+y)^2 = 19+2\sqrt{5} + 2i\sqrt{341} + 2\sqrt{5}-19$ $(x+y)^2 = (19-19) + (2\sqrt{5}+2\sqrt{5}) + 2i\sqrt{341}$ $(x+y)^2 = 4\sqrt{5} + 2i\sqrt{341}$
The value of $(x+y)^2$ is $\boxed{4\sqrt{5} + 2i\sqrt{341}}$ .

d) Calculate $\frac{x+y}{x-y}$
Express $\frac{x+y}{x-y}$ using the squared terms.
$\frac{x+y}{x-y} = \sqrt{\frac{(x+y)^2}{(x-y)^2}}$
Substitute the values of $(x+y)^2$ and $(x-y)^2$ .
$\frac{x+y}{x-y} = \sqrt{\frac{4\sqrt{5} + 2i\sqrt{341}}{4\sqrt{5} - 2i\sqrt{341}}}$
Rationalize the denominator inside the square root by multiplying the numerator and denominator by the conjugate of the denominator, which is $4\sqrt{5} + 2i\sqrt{341}$ .
$\frac{x+y}{x-y} = \sqrt{\frac{(4\sqrt{5} + 2i\sqrt{341})(4\sqrt{5} + 2i\sqrt{341})}{(4\sqrt{5} - 2i\sqrt{341})(4\sqrt{5} + 2i\sqrt{341})}}$ $\frac{x+y}{x-y} = \sqrt{\frac{(4\sqrt{5} + 2i\sqrt{341})^2}{(4\sqrt{5})^2 - (2i\sqrt{341})^2}}$
Calculate the denominator: $(4\sqrt{5})^2 - (2i\sqrt{341})^2 = (16 \cdot 5) - (4 \cdot i^2 \cdot 341) = 80 - (4 \cdot (-1) \cdot 341) = 80 - (-1364) = 80 + 1364 = 1444$ .
$\frac{x+y}{x-y} = \sqrt{\frac{(4\sqrt{5} + 2i\sqrt{341})^2}{1444}}$
Simplify the square root.
$\frac{x+y}{x-y} = \frac{4\sqrt{5} + 2i\sqrt{341}}{\sqrt{1444}}$
Since $\sqrt{1444} = 38$ :
$\frac{x+y}{x-y} = \frac{4\sqrt{5} + 2i\sqrt{341}}{38}$ $\frac{x+y}{x-y} = \frac{2(2\sqrt{5} + i\sqrt{341})}{38}$ $\frac{x+y}{x-y} = \frac{2\sqrt{5} + i\sqrt{341}}{19}$
The value of $\frac{x+y}{x-y}$ is $\boxed{\frac{2\sqrt{5} + i\sqrt{341}}{19}}$ .

:

a) Prove $(a+b)^2 - (a-b)^2 = 4ab$
Expand $(a+b)^2$ and $(a-b)^2$ .
$(a+b)^2 = a^2+2ab+b^2$ $(a-b)^2 = a^2-2ab+b^2$
Substitute the expanded forms into the expression.
$(a+b)^2 - (a-b)^2 = (a^2+2ab+b^2) - (a^2-2ab+b^2)$
Simplify the expression.
$= a^2+2ab+b^2 - a^2+2ab-b^2$ $= (a^2-a^2) + (b^2-b^2) + (2ab+2ab)$ $= 0 + 0 + 4ab$ $= 4ab$
Thus, $\boxed{(a+b)^2 - (a-b)^2 = 4ab}$ is proven.

b) Calculate $1000.001^2 - 999.999^2$
Use the identity $(a+b)^2 - (a-b)^2 = 4ab$ .
Let $a+b = 1000.001$ and $a-b = 999.999$ .
Solve for $a$ and $b$ .
Adding the two equations:
$(a+b) + (a-b) = 1000.001 + 999.999$
$2a = 2000$
$a = 1000$
Subtracting the second equation from the first:
$(a+b) - (a-b) = 1000.001 - 999.999$
$2b = 0.002$
$b = 0.001$
Substitute the values of $a$ and $b$ into the identity.
$1000.001^2 - 999.999^2 = 4 \cdot a \cdot b$ $= 4 \cdot (1000) \cdot (0.001)$
Calculate the result.
$= 4 \cdot (1000 \cdot \frac{1}{1000})$ $= 4 \cdot 1$ $= 4$
The value of $1000.001^2 - 999.999^2$ is $\boxed{4}$ .

The question is about simplifying algebraic expressions involving square roots, specifically exercises labeled '11' and '12' which include calculating xy, (x-y), (x+y) for given values of x and y, and proving algebraic identities like (a+b)2 − (a−b)2 = 4ab.

Step-by-step solution

Substitute the expressions for $x$ and $y$ .

Rewrite $\sqrt{2\sqrt{5}-19}$ using $i$ .

Substitute and multiply.

Apply the difference of squares formula $(a+b)(a-b) = a^2-b^2$ .

Expand $(x-y)^2$ .

Substitute $x^2$ , $y^2$ , and $xy$ .

Simplify the expression.

Expand $(x+y)^2$ .

Substitute $x^2$ , $y^2$ , and $xy$ .

Simplify the expression.

Express $\frac{x+y}{x-y}$ using the squared terms.

Substitute the values of $(x+y)^2$ and $(x-y)^2$ .

Rationalize the denominator inside the square root by multiplying the numerator and denominator by the conjugate of the denominator, which is $4\sqrt{5} + 2i\sqrt{341}$ .

Simplify the square root.

Expand $(a+b)^2$ and $(a-b)^2$ .

Substitute the expanded forms into the expression.

Simplify the expression.

Use the identity $(a+b)^2 - (a-b)^2 = 4ab$ .

Solve for $a$ and $b$ .

Substitute the values of $a$ and $b$ into the identity.

Calculate the result.

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More Mathematics Questions

The question is about simplifying algebraic expressions involving square roots, specifically exercises labeled '11' and '12' which include calculating xy, (x-y), (x+y) for given values of x and y, and proving algebraic identities like (a+b)2 − (a−b)2 = 4ab.

Step-by-step solution

Substitute the expressions for xxx and yyy.

Rewrite 25−19\sqrt{2\sqrt{5}-19}25​−19​ using iii.

Substitute and multiply.

Apply the difference of squares formula (a+b)(a−b)=a2−b2(a+b)(a-b) = a^2-b^2(a+b)(a−b)=a2−b2.

Expand (x−y)2(x-y)^2(x−y)2.

Substitute x2x^2x2, y2y^2y2, and xyxyxy.

Simplify the expression.

Expand (x+y)2(x+y)^2(x+y)2.

Substitute x2x^2x2, y2y^2y2, and xyxyxy.

Simplify the expression.

Express x+yx−y\frac{x+y}{x-y}x−yx+y​ using the squared terms.

Substitute the values of (x+y)2(x+y)^2(x+y)2 and (x−y)2(x-y)^2(x−y)2.

Rationalize the denominator inside the square root by multiplying the numerator and denominator by the conjugate of the denominator, which is 45+2i3414\sqrt{5} + 2i\sqrt{341}45​+2i341​.

Simplify the square root.

Expand (a+b)2(a+b)^2(a+b)2 and (a−b)2(a-b)^2(a−b)2.

Substitute the expanded forms into the expression.

Simplify the expression.

Use the identity (a+b)2−(a−b)2=4ab(a+b)^2 - (a-b)^2 = 4ab(a+b)2−(a−b)2=4ab.

Solve for aaa and bbb.

Substitute the values of aaa and bbb into the identity.

Calculate the result.

Still stuck on this one?

More Mathematics Questions

Substitute the expressions for $x$ and $y$ .

Rewrite $\sqrt{2\sqrt{5}-19}$ using $i$ .

Apply the difference of squares formula $(a+b)(a-b) = a^2-b^2$ .

Expand $(x-y)^2$ .

Substitute $x^2$ , $y^2$ , and $xy$ .

Expand $(x+y)^2$ .

Substitute $x^2$ , $y^2$ , and $xy$ .

Express $\frac{x+y}{x-y}$ using the squared terms.

Substitute the values of $(x+y)^2$ and $(x-y)^2$ .

Rationalize the denominator inside the square root by multiplying the numerator and denominator by the conjugate of the denominator, which is $4\sqrt{5} + 2i\sqrt{341}$ .

Expand $(a+b)^2$ and $(a-b)^2$ .

Use the identity $(a+b)^2 - (a-b)^2 = 4ab$ .

Solve for $a$ and $b$ .

Substitute the values of $a$ and $b$ into the identity.