Question

Please solve this!

5 points for each question! So, kindly give a complete answer, with all the steps to be mentioned!

Thank You :)​

Answer 1

ANSWER 12)

To Rationalise:

• 1/(√3 - √2 + 1)

Solution:

We are given that,

$\implies\dfrac{1}{\sqrt3-\sqrt2+1}$

$\implies\dfrac{1}{(\sqrt3-\sqrt2)+1}$

So,

$\implies\dfrac{1}{(\sqrt3-\sqrt2)+1}\times\dfrac{(\sqrt3-\sqrt2)-1}{(\sqrt3-\sqrt2)-1}$

Now,

$\implies\dfrac{(\sqrt3-\sqrt2)-1}{((\sqrt3-\sqrt2)+1)\times((\sqrt3-\sqrt2)-1)}$

As,

⇒ (a + b)(a - b) = a² - b²

So,

$\implies\dfrac{(\sqrt3-\sqrt2)-1}{(\sqrt3-\sqrt2)^2-(1)^2}$

As,

⇒ (a ± b)² = a² + b² ± 2ab

So,

$\implies\dfrac{(\sqrt3-\sqrt2)-1}{3+2-2\sqrt6-1}$

So,

$\implies\dfrac{\sqrt3-\sqrt2-1}{4-2\sqrt6}$

Now, we will rationalise it again,

$\implies\dfrac{\sqrt3-\sqrt2-1}{2(2-\sqrt6)}$

$\implies\dfrac{\sqrt3-\sqrt2-1}{2(2-\sqrt6)}\times\dfrac{2+\sqrt6}{2+\sqrt6}$

$\implies\dfrac{(\sqrt3-\sqrt2-1)(2+\sqrt6)}{2(2-\sqrt6)(2+\sqrt6)}$

$\implies\dfrac{2\sqrt3-2\sqrt2-2+3\sqrt2-2\sqrt3-\sqrt6}{2((2)^2-(\sqrt6)^2)}$

On regrouping,

$\implies\dfrac{(2\sqrt3-2\sqrt3)+(-2\sqrt2+3\sqrt2)-2-\sqrt6}{2((2)^2-(\sqrt6)^2)}$

So,

$\implies\dfrac{\sqrt2-\sqrt6-2}{2(4-6)}$

$\implies\dfrac{\sqrt2-\sqrt6-2}{2(-2)}$

$\implies\dfrac{\sqrt2-\sqrt6-2}{-4}$

Taking (-) common in the numerator,

$\implies\dfrac{-\!\!\!/\,(2+\sqrt6-\sqrt2)}{-\!\!\!/\,4}$

Hence,

$\implies\bf\dfrac{2+\sqrt6-\sqrt2}{4}$

ANSWER 13)

Given:

• √2 = 1.4
• √3 = 1.7

To Find:

1. 1/(√3-√2)
2. 1/(3+2√2)
3. (2-√3)/√3

Solution:

1) 1/(√3 - √2)

$\implies\dfrac{1}{\sqrt3-\sqrt2}$

On rationalising,

$\implies\dfrac{1}{\sqrt3-\sqrt2}\times\dfrac{\sqrt3+\sqrt2}{\sqrt3+\sqrt2}$

So,

$\implies\dfrac{\sqrt3+\sqrt2}{(\sqrt3)^2-(\sqrt2)^2}$

$\implies\dfrac{\sqrt3+\sqrt2}{3-2}$

So,

$\implies\bf\sqrt3+\sqrt2$

Substituting value of √3 as 1.7 & √2 as 1.4,

So,

$\implies(1.7)+(1.4)$

$\implies\bf3.1$

2) 1/(3 + 2√2)

$\implies\dfrac{1}{3+2\sqrt2}$

On rationalising,

$\implies\dfrac{1}{3+2\sqrt2}\times\dfrac{3-2\sqrt2}{3-2\sqrt2}$

So,

$\implies\dfrac{3-2\sqrt2}{(3)^2-(2\sqrt2)^2}$

$\implies\dfrac{3-2\sqrt2}{9-8}$

So,

$\implies\bf3-2\sqrt2$

Substituting value of √2 as 1.4,

So,

$\implies3-2(1.4)$

$\implies3-2.8$

$\implies\bf0.2$

3) (2 - √3)/√3

$\implies\dfrac{2-\sqrt3}{\sqrt3}$

On rationalising,

$\implies\dfrac{2-\sqrt3}{\sqrt3}\times\dfrac{\sqrt3}{\sqrt3}$

So,

$\implies\dfrac{(2-\sqrt3)\sqrt3}{(\sqrt3)^2}$

So,

$\implies\bf\dfrac{2\sqrt3-3}{3}$

Substituting value of √3 as 1.7,

So,

$\implies\dfrac{2(1.7)-3}{3}$

$\implies\dfrac{3.4-3}{3}$

$\implies\dfrac{0.4}{3}$

$\implies\bf0.1$

ANSWER 14)

To Evaluate:

• (4 - √5)/(4 + √5) + (4 + √5)/(4 - √5)

Solution:

We need to evaluate,

$\implies\dfrac{4-\sqrt5}{4+\sqrt5}+\dfrac{4+\sqrt5}{4-\sqrt5}$

Now, we will rationalise each expression simultaneously,

$\implies\left(\dfrac{4-\sqrt5}{4+\sqrt5}\times\dfrac{4-\sqrt5}{4-\sqrt5}\right)+\left(\dfrac{4+\sqrt5}{4-\sqrt5}\times\dfrac{4+\sqrt5}{4+\sqrt5}\right)$

So,

$\implies\dfrac{(4-\sqrt5)^2}{(4+\sqrt5)(4+\sqrt5)}+\dfrac{(4+\sqrt5)^2}{(4-\sqrt5)(4+\sqrt5)}$

$\implies\dfrac{16+5-8\sqrt5}{(4)^2-(\sqrt5)^2}+\dfrac{16+5+8\sqrt5}{(4)^2-(\sqrt5)^2}$

So,

$\implies\dfrac{21-8\sqrt5}{16-5}+\dfrac{21+8\sqrt5}{16-5}$

$\implies\dfrac{21-8\sqrt5}{11}+\dfrac{21+8\sqrt5}{11}$

$\implies\dfrac{21-8\sqrt5+21+8\sqrt5}{11}$

Hence,

$\implies\bf\dfrac{42}{11}$

ANSWER 15)

Given:

• x = (2 + √5)/(2 - √5)
• y = (2 - √5)/(2 + √5)

To Find:

• Value of: x² - y²

Solution:

We are given that,

$\implies x=\dfrac{2+\sqrt5}{2-\sqrt5}$

On rationalising,

$\implies x=\dfrac{2+\sqrt5}{2-\sqrt5}\times\dfrac{2+\sqrt5}{2+\sqrt5}$

So,

$\implies x=\dfrac{(2+\sqrt5)^2}{(2-\sqrt5)(2+\sqrt5)}$

$\implies x=\dfrac{4+5+4\sqrt5}{(2)^2-(\sqrt5)^2}$

$\implies x=\dfrac{9+4\sqrt5}{4-5}$

$\implies x=-(9+4\sqrt5)$

$\implies x=-9-4\sqrt5$

We are also given that,

$\implies y=\dfrac{2-\sqrt5}{2+\sqrt5}$

On rationalising,

$\implies y=\dfrac{2-\sqrt5}{2+\sqrt5}\times\dfrac{2-\sqrt5}{2-\sqrt5}$

So,

$\implies y=\dfrac{(2-\sqrt5)^2}{(2-\sqrt5)(2+\sqrt5)}$

$\implies y=\dfrac{4+5-4\sqrt5}{(2)^2-(\sqrt5)^2}$

$\implies y=\dfrac{9-4\sqrt5}{4-5}$

$\implies y=-(9-4\sqrt5)$

$\implies y=-9+4\sqrt5$

Now, we need to find, x² - y²

We can also write it as,

$\implies x^2-y^2=(x+y)(x-y)$

Substituting the value of x and y,

$\implies \bigg((-9-4\sqrt5)+(-9+4\sqrt5)\bigg)\times\bigg((-9-4\sqrt5)-(-9+4\sqrt5)\bigg)$

$\implies \bigg(-9-4\sqrt5-9+4\sqrt5\bigg)\times\bigg(-9-4\sqrt5+9-4\sqrt5\bigg)$

So,

$\implies(-18)\times(-8\sqrt5)$

$\implies (18\times8)\sqrt5$

$\implies\bf144\sqrt5$

Answer 2

The curve C has equation

y= x3 + ax2 + bx + 8

where a and b are constants.

The gradient of C at the point (2,6) is 9.8

Find the value of a and the value of b.

Show clear algebraic working.