Let f(x) = ax² + bx + c be the polynomial satisfying the given conditions.
f(0) = a(0)² + b(0) + c = 1, implies that c = 1. Now the other two conditions f(-2) = 0; f(1) = 0 give 4a - 2b + c = 0 and a + b + c = 0.
Using c = 1, we get 4a - 2b = -1 and a + b = -1. Solving these two equations we get a = b = \(-\frac{1}{2}\) and thus, we have f(x) =\(-\frac{1}{2}x^2-\frac{1}{2}x+1\)

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  \({{1}\over{2}}\) log₁₀ \((11+4\sqrt{7})\) = log₁₀ (2 + x).
\(\Rightarrow\) \({{1}\over{2}} \) log₁₀ \((7+4+4\sqrt{7})\) = log₁₀(2 + x)
\(\Rightarrow \) \({{1}\over{2}}\) log₁₀ \(((\sqrt{7})^2+2^2+2.2\sqrt{7})\) = log₁₀(2 + x)
\(\Rightarrow\)  \({{1}\over{2}}\) log₁₀\({(\sqrt{7}+2)}^{2}\)= log₁₀(2 + x)
\(\Rightarrow\)  log₁₀\((\sqrt{7}+2)\) = log₁₀(2 + x)
\(\Rightarrow\)  \((\sqrt{7}+2)\)= (2 + x)
\(\Rightarrow\)   \(x=\sqrt{7}\)

Given \(log_2\left({{\sqrt [ 3 ]{4 } }\over{4^2\sqrt{8}}} \right)\)
= \({log}_{2}\sqrt [ 3 ]{4 }-{log}_{2}4^2(\sqrt{8})\)
= \(log_24^{1/3}-[log_24^2+log_2\sqrt{8}]\)
= log²(2²)^1/3- log²(2²)²- log²(2³)^1/2
= log²2^1/3- log²2⁴- log²2^3/2
\(={{2}\over{3}}(1)-4(1)-{{3}\over{2}}(1)\)   \([\because {log}^{2}_{2}=1]\)
\(={{4-24-9}\over{6}}={{-29}\over{6}}\)

Let the output two years ago be x units and annual increase = r%
\(\therefore\)  Output in the last year \(=x\left( {{100+r}\over{100}}\right) \) units.   \(\left[ I={{PNR}\over{100}} \right]\)
and output in the present year.
\(=x\left({{100+r}\over{100}} \right)^2\)
\(\therefore\) By the given conduction, \(x\left({{100+r}\over{1000}} \right)=2x\)
\(\Rightarrow\)    \(\left({{100+9r}\over{100}} \right)^2=2\)
\(\Rightarrow\)  (100 + r)² = 20,000
\(\Rightarrow\)  100² + r² + 200r - 20,000 = 0
\(\Rightarrow\) r²+ 200r - 10,000 = 0
\(\Rightarrow\)   \(r={{-200\pm\sqrt{40,000+40,000}}\over{2}}=r={{-200\pm\sqrt{80,000}}\over{2}}\)
\(\Rightarrow\)   \(r={{-200\pm200\sqrt{2}}\over{2}}\)
\(\Rightarrow\)   \(r={{2(-100\pm100\sqrt{2})}\over{2}}=-100\pm100\sqrt{2}\)
\(\Rightarrow\)   \(r=100(\sqrt{2}-1)\) or \(100(-1-\sqrt{2})\)
\(\Rightarrow\)   \(r=100(\sqrt{2}-1)\)      \([\because r > 0]\)
\(\Rightarrow\)   \(r=100(1.141-1)=41.42\%\)

Given quadratic equation is \(\sqrt [ 8 ]{{{x}\over{x+3}} } -\sqrt{{{x+3}\over{x}}}=2.\)
Let y \(=\sqrt{x\over x+3}\Rightarrow{1\over y}=\sqrt{x+3\over x}\)
∴ (1) becomes 8y - \(\frac{1}{y}\) = 2

\(⇒\ {8y^2-1\over y}=2⇒8y^2-1=2\)
8y²-2y-1 = 0
(2y - 1)(4y + 1) = 0
2y = 1 or 4y = -1
\(y={1\over 2}\)or \(y={-1\over 4}\)
\(\sqrt{x\over x+3}={1\over 2}\sqrt{x\over x+3}={1\over 2}\ or\ \sqrt{x\over x+3}={-1\over 4}\)
Case(i) \(\sqrt{x\over x+3}={1\over 2}\Rightarrow{x\over x+3}={-1\over 4}\)
4x = x + 3 ⇒ 3x = 3 ⇒ x = 1
Case(ii) \(\sqrt{x\over x+3}={-1\over 4}\)
This is impossible since LHS is non-negative.
∴ The root is 1.

Given quadratic equation is
5^2x-5^x+3 + 125 = 5x
\(\Rightarrow\) (5^x)²-5^x 5³- 5^x + 125 = 0
\(\Rightarrow\) (5^x)²-125 5^x- 5^x + 125 = 0
\(\Rightarrow\) (5^x)²-126.5^x+125=0
Let 5^x=y
\(\Rightarrow\)y²-126y + 125 = 0
\(\Rightarrow\)(y-1)(y-125) = 0
\(\Rightarrow\)    y = 1 or 125
\(\Rightarrow\)   5^x = 1 or 125
Case (i) When 5^x = 1 \(\Rightarrow\) 5⁰ \(\Rightarrow\) x = 0
Case (ii) When 5^x = 125 \(\Rightarrow\) 5^x = 5³\(\Rightarrow\) x = 3
\(\therefore\) The roots are 0, 3.

Given 23x<100.
(i) When x is a natural number
23x < 100
\(\Rightarrow\) x < 100/23
\(\Rightarrow\) x < 4.347
\(\Rightarrow\) x = 1,2,3
\(\therefore\) Solution set is { 1, 2, 3, 4}
(ii) When x is an integer.
x < 4.347
\(\Rightarrow\) x = { ...... -4, -3, -2, -1, 0, 1, 2, 3 }
Hence solution set is { ...... -4, -3, -2, -1, 0, 1, 2, 3 }

Let the number of hours to complete the job is x.
Wages from the first scheme = 500 + 70x
Wages from the second scheme = 120x
Given 500 + 70x > 120x
\(\Rightarrow\)  500 > 120x -70x
\(\Rightarrow\)  500 > 50x
       
\(\Rightarrow\) 10 > x
\(\Rightarrow\) x < 10
\(\therefore\) The value of x so that the first scheme gives better wages is x = 1, 2, 3, 4, 5, 6, 7, 8, 9.

LHS = \(log2+16log{16\over 15}+12log{25\over 24}+7log{81\over 80}\)
\(=log2+log\left(16\over 15\right)^{16}+log\left(25\over 24\right)^{12}+log \left(81\over 80\right)^7\)
\(=log2\times{(2^4)^{16}\over (3\times5)^{16}}\times{(5^2)^{12}\over (2^2\times3)^{12}}\times{(3^4)^7\over 2^{28}\times5^7}\)
\(=log2^1\times{2^{64}\over 3^{16}}\times{5^{24}\over 2^{36}\times3^{12}}\times{3^{28}\over 2^{28}\times5^7}\)
\(=log{2^{1+64}.5^{24}.3^{28}\over 3^{16+12}.5^{16+7}.2^{36+28}}\)  \(\left[∵\ {a^m\over a^n}=a^{m-n} \right]\)

= log 2^65-64 x 5^24-23 = log 2¹ \(\times\) 5¹ = log₁₀¹⁰ = 1 = RHS