Please solve q4 and 5?

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Please solve q4 and 5?

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Answer 1 · 2018-02-17T17:23:26

$n = 0$ $n=0$

Explanation:

Question 4:
Given:

$n = \sqrt{6 + \sqrt{11}} + \sqrt{6 - \sqrt{11}} - \sqrt{22}$ $n=sqrt(6+sqrt11)+sqrt(6-sqrt11)-sqrt22$

Let,
$\sqrt{6 + \sqrt{11}} = \sqrt{p} + \sqrt{q}$ $sqrt(6+sqrt11)=sqrtp+sqrtq$
Then,
$\sqrt{6 - \sqrt{11}} = \sqrt{p} - \sqrt{q}$ $sqrt(6-sqrt11)=sqrtp-sqrtq$
Squaringand adding

$(6 + \sqrt{11}) + (6 - \sqrt{11}) = p + q + 2 \sqrt{p q} + p + q - 2 \sqrt{p q}$ $(6+sqrt11)+(6-sqrt11)=p+q+2sqrt(pq)+p+q-2sqrt(pq)$
$12 = 2 (p + q)$ $12=2(p+q)$
$p + q = \frac{12}{2} = 6$ $p+q=12/2=6$
$p + q = 6$ $p+q=6$

Squaring and subtracting

$(6 + \sqrt{11}) - (6 - \sqrt{11}) = (p + q + 2 \sqrt{p q}) - (p + q - 2 \sqrt{p q})$ $(6+sqrt11)-(6-sqrt11)=(p+q+2sqrt(pq))-(p+q-2sqrt(pq))$ =
$2 \sqrt{11} = 4 \sqrt{p q}$ $2sqrt11=4sqrt(pq)$
$\sqrt{p q} = \frac{2 \sqrt{11}}{4} = \frac{\sqrt{11}}{2}$ $sqrt(pq)=(2sqrt11)/4=sqrt(11)/2$

Squaring
$p q = \frac{11}{4} = 2.75$ $pq=11/4=2.75$

$x^{2} - S u m x + P r o d u c t = 0$ $x^2-Sumx+Product=0$

$x^{2} - 6 x + 2.75 = 0$ $x^2-6x+2.75=0$
$x^{2} - 5.5 x - 0.5 x + 2.75 = 0$ $x^2-5.5x-0.5x+2.75=0$

$x (x - 5.5) - 0.5 (x - 5.5) = 0$ $x(x-5.5)-0.5(x-5.5)=0$

$(x - 5.5) (x - 0.5) = 0$ $(x-5.5)(x-0.5)=0$

$x - 5.5 = 0 \to x = 5.5$ $x-5.5=0tox=5.5$
$x - 0.5 = 0 \to x = 0.5$ $x-0.5=0tox=0.5$
One of the roots can be p, other will be q.
Thus,

$\sqrt{6 + \sqrt{11}} = \sqrt{5.5} + \sqrt{0.5}$ $sqrt(6+sqrt11)=sqrt5.5+sqrt0.5$

It follows that
$\sqrt{6 - \sqrt{11}} = \sqrt{5.5} - \sqrt{0.5}$ $sqrt(6-sqrt11)=sqrt5.5-sqrt0.5$

Now,
$\sqrt{6 + \sqrt{11}} + \sqrt{6 - \sqrt{11}} - \sqrt{22} = \sqrt{5.5} + \sqrt{0.5} + \sqrt{5.5} - \sqrt{0.5} - \sqrt{22}$ $sqrt(6+sqrt11)+sqrt(6-sqrt11)-sqrt22=sqrt5.5+sqrt0.5+sqrt5.5-sqrt0.5-sqrt22$
$= 2 \sqrt{5.5} - \sqrt{22}$ $=2sqrt5.5-sqrt22$
$= q r t 4 \sqrt{5.5} = \sqrt{22}$ $=qrt4sqrt5.5=sqrt22$
$= \sqrt{4 \times 5.5} - \sqrt{22}$ $=sqrt(4xx5.5)-sqrt22$
$= \sqrt{22} - \sqrt{22}$ $=sqrt22-sqrt22$
$= 0$ $=0$
Thus,

$n = 0$ $n=0$

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Please solve q4 and 5?

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