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\(9^{y} - 4 \times 3^{y} + 3 = 0\) \(\equiv (3^{2})^{y} - 4 \times 3^{y} + 3 = 0\) \((3^{y})^{2} - 4 \times 3^{y} + 3 = 0\) Let \(3^{y}\) be r. Then, \(r^{2} - 4r + 3 = 0\) Solving the equation, \(r^{2} - 3r - r + 3 = 0\) \(r(r - 3) - 1(r - 3) = 0\) \((r - 3)(r - 1) = 0\) \(\therefore \text{r = 3 or 1}\) Recall, \(3^{y} = r\) \(3^{y} = 3 = 3^{1} \text{ or } 3^{y} = 1 = 3^{0}\) \(\implies \text{y = 1 or 0}\)

S = \(\sqrt{\frac{2R + T}{2RT}}\) Squaring both sides, \(S^{2} = \frac{2R + T}{2RT}\) \(S^{2} (2RT) = 2R + T\) \(2S^{2} RT - 2R = T\) \(R = \frac{T}{2TS^{2} - 2}\) = \(\frac{T}{2(TS^2 - 1)}\)

where C = cost, S = number of students, a and K are constants. C = a + kS. If C = 74, S = 20 C = 96, S = 30, C= ? S = 15 74 = a + 20k......(1) 96 = a + 30k......(2) subtract equation (1) from (2) 96 = a + 30k - 74 = a + 20k -------------- 22 = 10k k = 2.2 find a 74 = a + 44 a = 30 C = 30 + 2.2S when S = 15, C = 30 + 2.2 x 15 = 30 + 33 = N63

\(f(x) = 2x^2 - 5x + 3\) \(f(x + 1) = 2(x + 1)^2 - 5(x + 1) + 3\) = \(2(x^2 + 2x + 1) - 5x - 5 + 3\) = \(2x^2 + 4x + 2 - 5x - 2\) = \(2x^2 - x\)

\(2^{(x^3 - x^2 - 2x)} = 1\) \(x^3 - x^2 - 2x = 0\) \(x(x^2 - x - 2) = 0\) \(x^2 - x - 2 = 0\) \((x + 1)(x - 2) = 0\) x = 2 is the positive answer.

\(\frac{324 - 4x^2}{2x + 18}\) = \(\frac{18^2 - (2x)^2}{2x + 18}\) = \(\frac{(18 - 2x)(18 + 2x)}{(2x + 18)}\) 18 - 2x = 2(9 - x) or -2(x - 9)

\(y^3 -4xy + xy^3 - 4y \) = \(y^3(1 + x) - 4y(1 + x) \((y^3 - 4y)(1 + x) = (y^3(1 + x) - 4y(1 + x))\) ∴ = \(y(1 + x)(y + 2)(y - 2)\)

4a2 - 12ab - C2 + 9b2 rearranges: (4a2 - 12ab + 9b2) - c2 (2a - 3b)(2a - 3b) - c2 = (2a - 3b)2 - c2 = (2a - 3b + c)(2a - 3b - c)

\(\frac{1}{2}\)(3y - 4x)2 = (8x2 + kxy + Ly2) \(\frac{1}{2}\)(9y2 - 24xy + 16x2) = 8x2 + kxy + Ly2 \(\frac{9}{2}\)y2 - 12xy) = kxy + Ly2 k = -12 ∴ L = \(\frac{9}{2}\)

\(2x^{-1} - 3y^{-1} = 4; 4x^{-1} + y^{-1} = 1\) Let \(x^{-1}\) = a and \(y^{-1}\)= b 2a - 3b = 4 .......(i) 4a + b = 1 .........(ii) (i) x 3 = 12a + 3b = 3........(iii) 2a - 3b = 4 ...........(i) (i) + (iii) = 14a = 7 ∴ a = \(\frac{7}{14}\) = \(\frac{1}{2}\) From (i), 3b = 2a - 4 3b = 1 - 4 3b = -3 ∴ b = -1 From substituting, \(2^{-1} = x^{-1}\) ∴ x = 2 \(y^{-1} = -1, y = -1\)

4x2 + 5x + Q To make a complete square, the coefficient of x2 must be 1 = x2 + \(\frac{5x}{4}\) + \(\frac{Q}{4}\) Then (half the coefficient of x2) should be added i.e. x2 + \(\frac{5x}{4}\) + \(\frac{25}{64}\) ∴ \(\frac{Q}{4}\) = \(\frac{25}{64}\) Q = \(\frac{4 \times 25}{64}\) = \(\frac{25}{16}\)

\(\frac{r}{a}\) + \(\frac{r}{b}\) + \(\frac{r}{c}\) > 1 = \(\frac{bcr + acr + abr}{abc}\) > 1 r(bc + ac + ba > abc) = r > \(\frac{abc}{bc + ac + ab}\)

\(\frac{1}{x + 1}\) - \(\frac{1}{x - 2}\) = \(\frac{x - 2 - x - 1}{(x + 1)(x - 2)}\) = \(\frac{-3}{(x + 1)(2 - x)}\)

\[\frac{x(x + 1)^{-\frac{1}{2}} - (x + 1)^{\frac{1}{2}}}{(x + 1)^{\frac{1}{2}}}\] The numerator can be rewritten as: \[x(x + 1)^{-\frac{1}{2}} - (x + 1)^{\frac{1}{2}} = \frac{x}{(x + 1)^{\frac{1}{2}}} - (x + 1)^{\frac{1}{2}}\] Combining the terms in the numerator over a common denominator: \[= \frac{x - (x + 1)}{(x + 1)^{\frac{1}{2}}}\] \[= \frac{x - x - 1}{(x + 1)^{\frac{1}{2}}} = \frac{-1}{(x + 1)^{\frac{1}{2}}}\] Substituting this back into the original expression gives: \[\frac{\frac{-1}{(x + 1)^{\frac{1}{2}}}}{(x + 1)^{\frac{1}{2}}}\] This simplifies to: \[\frac{-1}{(x + 1)^{\frac{1}{2}} \cdot (x + 1)^{\frac{1}{2}}} = \frac{-1}{(x + 1)}\] = \[{\frac{-1}{x + 1}}\]

Sum of nth term of a G.P = Sn = \(\frac{ar^n - 1}{r - 1}\) sum of the first two terms = \(\frac{ar^2 - 1}{r - 1}\) x = a(r + 1) sum of the last two terms = Sn - Sn - 2 = \(\frac{ar^n - 1}{r - 1}\) - \(\frac{(ar^{n - 1})}{r - 1}\) = \(\frac{a(r^n - 1 - r^{n - 2} + 1)}{r - 1}\) (r2 - 1) ∴ \(\frac{ar^{n - 2}(r + 1)(r - 1)}{1}\)= arn - 2(r + 1) = y = a(r + 1)r^n - 2 y = xrn - 2 = yrn - 2 \(\frac{y}{x}\) = r = (\(\frac{y}{x}\))\(\frac{1}{n - 2}\)