1. Problem 1: Given that $25x^2 - 20x + k$ is a perfect square. A quadratic is a perfect square exactly when its discriminant is zero. Compute the discriminant: $$\Delta = (-20)^2 - 4\cdot25\cdot k = 400 - 100k.$$ Set the discriminant to zero and solve: $$400 - 100k = 0 \Rightarrow k = 4.$$ Answer: $k = 4$. 2. Problem 2: Simplify $\dfrac{2y^2 - xy - x^2}{2x^2 - 2y^2}$. Factor the numerator as a quadratic in $y$: $2y^2 - xy - x^2 = (2y + x)(y - x)$. Factor the denominator: $2x^2 - 2y^2 = 2(x^2 - y^2) = 2(x - y)(x + y)$. Use $y - x = -(x - y)$ to cancel common factors and simplify: $$\dfrac{2y^2 - xy - x^2}{2x^2 - 2y^2} = -\dfrac{2y + x}{2(x + y)}.$$ Note (useful identity): $a^2 - b^2 = (a - b)(a + b)$. Answer: $-\dfrac{2y + x}{2(x + y)}$. 3. Problem 3: Solve $\dfrac{23}{x} - \dfrac{1}{x^2} - 120 = 0$, give answers to 3 d.p. Multiply both sides by $x^2$ (assuming $x\neq 0$) to obtain a quadratic: $$23x - 1 - 120x^2 = 0 \Rightarrow 120x^2 - 23x + 1 = 0.$$ Apply the quadratic formula: $$x = \dfrac{23 \pm \sqrt{23^2 - 4\cdot120\cdot1}}{240} = \dfrac{23 \pm \sqrt{49}}{240}.$$ Thus $$x = \dfrac{23 \pm 7}{240}$$ which gives $$x = \dfrac{30}{240} = \dfrac{1}{8} = 0.125$$ and $$x = \dfrac{16}{240} = \dfrac{1}{15} \approx 0.067.$$ Answer (3 d.p.): $0.125$ and $0.067$. 4. Problem 4: Simplify $$\dfrac{16x^2 - 4}{4x^2 + 2x - 2} \div \dfrac{2x - 2}{x + 1}.$$ Replace division by multiplication by the reciprocal: $$\dfrac{16x^2 - 4}{4x^2 + 2x - 2} \cdot \dfrac{x + 1}{2x - 2}.$$ Factor each polynomial: $16x^2 - 4 = 4(4x^2 - 1) = 4(2x - 1)(2x + 1)$, and $4x^2 + 2x - 2 = 2(2x^2 + x - 1) = 2(2x - 1)(x + 1)$, and $2x - 2 = 2(x - 1)$. Cancel common factors $(2x - 1)$ and $(x + 1)$ to get $$\dfrac{4(2x + 1)}{2\cdot 2(x - 1)} = \dfrac{2x + 1}{x - 1}.$$ Answer: $\dfrac{2x + 1}{x - 1}$. 5. Problem 5: Simplify $$\dfrac{(4x + 2y)^2 - (2y - 4x)^2}{(2z + y)^2 - (y - 2x)^2}.$$ Use the difference of squares $A^2 - B^2 = (A - B)(A + B)$ with $A = 4x + 2y$ and $B = 2y - 4x$. Compute $A - B = (4x + 2y) - (2y - 4x) = 8x$ and $A + B = (4x + 2y) + (2y - 4x) = 4y$, so the numerator is $8x\cdot 4y = 32xy$. For the denominator let $C = 2z + y$ and $D = y - 2x$. Then $C - D = 2z + y - (y - 2x) = 2(z + x)$ and $C + D = 2z + y + y - 2x = 2(z + y - x)$, so the denominator is $2(z + x)\cdot 2(z + y - x) = 4(x + z)(z + y - x)$. Therefore the fraction simplifies to $$\dfrac{32xy}{4(x + z)(z + y - x)} = \dfrac{8xy}{(x + z)(z + y - x)}.$$ Answer: $\dfrac{8xy}{(x + z)(z + y - x)}$. 6. Problem 6: In triangle $ABC$ with angle $B = 90^\circ$, given $AB = 3x + 5$, $BC = \sqrt{611}$ and hypotenuse $AC = 7x + 2$, find $x$ and hence the area. By Pythagoras: $$AB^2 + BC^2 = AC^2$$ so $$ (3x + 5)^2 + (\sqrt{611})^2 = (7x + 2)^2.$$ Compute and simplify: $9x^2 + 30x + 25 + 611 = 49x^2 + 28x + 4$, which gives $$40x^2 - 2x - 632 = 0$$ and dividing by 2 yields $$20x^2 - x - 316 = 0.$$ Quadratic formula: $$x = \dfrac{1 \pm \sqrt{1 + 4\cdot20\cdot316}}{40} = \dfrac{1 \pm \sqrt{25281}}{40} = \dfrac{1 \pm 159}{40}.$$ The positive solution is $$x = \dfrac{160}{40} = 4$$ (the negative root gives negative side lengths and is rejected). Compute sides: $AB = 3x + 5 = 17$, $BC = \sqrt{611}$, so the area is $$\dfrac{1}{2}AB\cdot BC = \dfrac{1}{2}\cdot 17 \cdot \sqrt{611} = \dfrac{17}{2}\sqrt{611}.$$ Numeric approximation (optional): $\dfrac{17}{2}\sqrt{611} \approx 210.093$. Answer: $x = 4$, area $= \dfrac{17}{2}\sqrt{611}$. 7. Problem 7: Solve the compound inequality $7x - 4 \le 9x + 2 < 3x + 14$, represent on a number line and state integral values. Split into two inequalities: (i) $7x - 4 \le 9x + 2$ and (ii) $9x + 2 < 3x + 14$. Solve (i): $7x - 4 \le 9x + 2 \Rightarrow -6 \le 2x \Rightarrow x \ge -3$. Solve (ii): $9x + 2 < 3x + 14 \Rightarrow 6x < 12 \Rightarrow x < 2$. Combine the two: $$-3 \le x < 2,$$ which on a number line is the interval $[-3,2)$. The integral values satisfying the compound inequality are $-3, -2, -1, 0, 1$. Answer: solution set $[-3,2)$ and integral values $-3,-2,-1,0,1$.