Observe that \(\sqrt{x+20}\) is defined only if x + 20 ≥ 0
By definition, \(\sqrt{x+20}\ge0\). So, x is positive.
Now squaring we get x²= x + 20. x²- x - 20 = 0
(x - 5)(x + 4) = 0, which gives x = 5, x = -4
Since, x is positive, the required solution is x = 5

The given equation is equivalent to the system (x + 4) ≥ 0 and 6 - 4x - x² = (x + 4)²
This implies x ≥ -4 and x² + 6x + 5 = 0. Thus x = -1, -5.
But only x = -1 satisfies both the conditions. Hence, x = -1.

Given a²+ b² = 7ab
Adding 2ab both sides we get,
a²+b²+2ab = 7ab + 2ab
⇒ (a+b)² = 9ab
⇒ \(\frac { (a+b)^{ 2 } }{ 9 } \)  = ab
⇒  \(\left( \frac { a+b^{ 2 } }{ 9 } \right) ^{ 2 }\)  = ab
Taking square root,we get
\(\frac { a+b }{ 3 } \)  = ab
\(log\left( \frac { a+b }{ 3 } \right) =log(ab)^{ \frac { 1 }{ 2 } }\)
= \(\frac { 1 }{ 2 } \) log (ab)
log \(\left( \frac { a+b }{ 3 } \right) \) = \(\frac { 1 }{ 2 } \) [log a + log b]
Hence proved.

Given log₈ x + log₄ x + log₂ x = 11
⇒ \(\frac { 1 }{ { log }_{ x }^{ 8 } } +\frac { 1 }{ { log }_{ x }^{ 4 } } +\frac { 1 }{ { log }_{ x }^{ 2 } } \) = 11
⇒ \(1-\frac { 1 }{ { { log }_{ x }^{ { 2 }^{ 3 } } } } +\frac { 1 }{ { log }_{ x }^{ { 2 }^{ 2 } } } +\frac { 1 }{ { log }_{ x }^{ 2 } } \)  = 11
⇒  \(\frac { 1 }{ 3{ { log }_{ x }^{ 2 } } } +\frac { 4 }{ 2{ { log }_{ x }^{ 2 } } } +\frac { 1 }{ { log }_{ x }^{ 2 } } \) = 11
⇒ \(\frac { 1 }{ { { log }_{ x }^{ 2 } } } \left( \frac { 1 }{ 3 } +\frac { 1 }{ 2 } +1 \right) \)  = 11
⇒   \(\frac { 1 }{ { { log }_{ x }^{ 2 } } } \left( \frac { 2+3+6 }{ 6 } \right) \)= 11
⇒ \(\frac { 1 }{ { { log }_{ x }^{ 2 } } } \left( \frac { 11 }{ 6 } \right) \) = 11
⇒   \(\frac { 1 }{ { { log }_{ x }^{ 2 } } } =11\times \frac { 6 }{ 11 } =6\)
⇒  \(\frac { 1 }{ { { log }_{ x }^{ 2 } } } \) = 6
⇒  log₂^x = 6 
⇒ 2⁶ = x
⇒ x = 64
∴ x = 64

The means \(-{{1}\over{4}}-3\le{3\over4}x\le{1\over 4}.\)
\(⇒-{{1}\over{4}}-3≤-{3\over4}x≤{1\over4}-3\)
\(⇒-{13\over 4}≤-{3\over4}x≤-{11\over4}\)
Multiplying by 4 throughout we get,
\(-13≤-3x≤-11\)
\({-13\over -3}≥x≥{-11\over3}.\)
\(\therefore\) The Solution set is \(\left[ \frac { -11 }{ 3 } ,\frac { 13 }{ 3 } \right] \)

Let the two irrational numbers be \(5+\sqrt { 7 } \) and \(7-\sqrt { 7 } \)
Their sum = \(\left( 5+\sqrt { 7 } \right) +\left( 7-\sqrt { 7 } \right) \ = \ 5+\sqrt { 7 } +7-\sqrt { 7 } \)
= 5 + 7 = 12 which is a rational number.
Consider  the two irrational numbers \(4+\sqrt { 6 } \) and \(4-\sqrt { 6 } \)
Their product = \(\left( 4+\sqrt { 6 } \right) +\left( 4-\sqrt { 6 } \right) ={ 4 }^{ 2 }-{ \left( \sqrt { 6 } \right) }^{ 2 }\)= 16 - 6 = 10 which is a rational number.

Suppose √3 is a rational number
Then √3 can be written as \(√3={m\over n}\)
Where m and n are rational numbers with no common factors other than 1.
Squaring both sides we get,
\(3={m^2\over n^2}⇒3n^2=m^2\)
multiplying by 2 we get
6n² = 2m² ⇒ 3(2n²) = 2 m²
Since 2n² is divisible by 2, m² is also an even number
⇒ m must be even
⇒ m = 2k for some natural number k
⇒ 3n² = (2k)²
⇒ 3n² = 4k²
⇒ n is also an even number
Thus both m and n are even numbers having a common factor.
This contradicts our initial assumption that m and n do not have a common factor
Hence √3 cannot be a rational number.
⇒ √3 is an irrational number.
Hence proved.

Given \(\left( { x }^{ +\frac { 1 }{ 2 } }+{ x }^{ -\frac { 1 }{ 2 } } \right) ^{ 2 }=\frac { 9 }{ 2 } \)
⇒ \(x+\frac { 1 }{ 2 } +{ 2x }^{ \frac { 1 }{ 2 } }.\frac { 1 }{ { x }^{ -\frac { 1 }{ 2 } } } =\frac { 9 }{ 2 } \)
⇒  \(x+\frac { 1 }{ x } +2=\frac { 9 }{ 2 } \)
⇒ \(x+\frac { 1 }{ x } =\frac { 9 }{ 2 } -2=\frac { 9-4 }{ 2 } =\frac { 5 }{ 2 } \)
Consider \(\left( { x }^{ \frac { 1 }{ 2 } }-{ x }^{ -\frac { 1 }{ 2 } } \right) ^{ 2 }\)
= \(x+\frac { 1 }{ x } -2x^{ \frac { 1 }{ 2 } }.\frac { 1 }{ { x }^{ \frac { 1 }{ 2 } } } \)
= \(x+\frac { 1 }{ x } -2\)    [From(1)]
=   \(\frac { 5 }{ 2 } -2=\frac { 5-4 }{ 2 } =\frac { 1 }{ 2 } \)       [using 1]
∴  \(x^{ \frac { 1 }{ 2 } }-x^{ -\frac { 1 }{ 2 } }=\pm \frac { 1 }{ \sqrt { 2 } } \)
⇒ \(x^{ \frac { 1 }{ 2 } }-x^{ -\frac { 1 }{ 2 } }=\frac { 1 }{ \sqrt { 2 } } \)   since x > 1

First we consider equation x = 2. It is a line parallel to y axis at a distance of 2 units from it. This line divides the cartesian plane into two parts. Substituting (0, 0) in the inequality we get 0 ≥ 2 which is false. Hence the region which does not contain the origin is represented by the inequality x ≥ 2. The shaded region is the required solution set of the given inequality. Since x ≥ 2, the points on the line x = 2 are also solutions.

If 3x + 5y = 45

All points bounded above x = 0, y = 0 and 3x + 5y = 45 is required region. Darkly shaded area will represents the solution set of the given linear inequalities.