1. **Problem:** GOAL and FCTR are squares. The area of GOAL is $9x^2 - 12x + 4$ cm$^2$ and is larger than the area of FCTR by $5x^2 + 8x - 21$ cm$^2$. Find the length of a side of FCTR. 2. The area of FCTR is $$\text{area}_{GOAL} - \text{difference} = (9x^2 - 12x + 4) - (5x^2 + 8x - 21)$$ 3. Simplify the expression: $$9x^2 - 12x + 4 - 5x^2 - 8x + 21 = (9x^2 - 5x^2) + (-12x - 8x) + (4 + 21) = 4x^2 - 20x + 25$$ 4. Since FCTR is a square, its area equals side$^2$. Let $s$ be the side length: $$s^2 = 4x^2 - 20x + 25$$ 5. Factor the quadratic: $$4x^2 - 20x + 25 = (2x - 5)^2$$ 6. Therefore, the length of a side of FCTR is: $$s = |2x - 5|$$ --- 4. **Problem:** The volume of a rectangular prism is $343 m^3 - 8$ dm$^3$. Find the area of its base. 5. Note units: $1 m^3 = 1000 dm^3$. Convert $343 m^3$ to dm$^3$: $$343 m^3 = 343 \times 1000 = 343000 dm^3$$ 6. Volume is: $$343000 - 8 = 342992 \, dm^3$$ 7. Recognize $343000 - 8 = (70^3) - 2^3$ (since $70^3=343000$, $2^3=8$). 8. Use difference of cubes formula: $$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$$ 9. So, $$342992 = (70 - 2)(70^2 + 70*2 + 2^2) = 68 (4900 + 140 + 4) = 68 \times 5044$$ 10. Volume $= \text{base area} \times \text{height}$, so if height = 68, $$\text{base area} = 5044 \, \text{dm}^2$$ --- 5. **Problem:** A square cardboard with area $729w^4$ cm$^2$ has a smaller square of area $729w^2$ cm$^2$ cut out. Find the dimensions of the rectangle formed by the remaining pieces. 6. Side length of large square: $$\sqrt{729w^4} = 27w^2$$ 7. Side length of smaller square: $$\sqrt{729w^2} = 27w$$ 8. After removing the smaller square, the remaining pieces form a rectangle with dimensions: - Length: $27w^2$ - Width: $27w^2 - 27w = 27w(w - 1)$ --- 6. **Problem:** Parallelogram area is $y^2 - y - 156$ m$^2$. Altitude is 25 m less than base. Find the altitude. 7. Let base = $b$, altitude = $b - 25$ 8. Area formula: $$\text{area} = b(b - 25) = b^2 - 25b = y^2 - y -156$$ 9. Equate: $$b^2 - 25b - (y^2 - y - 156) = 0 \implies b^2 - 25b - y^2 + y +156 = 0$$ (Since $y$ is variable, assume $b$ corresponds to $y$; or interpret as quadratic in $b$.) 10. Instead, factor area: $$y^2 - y -156 = (y - 13)(y + 12)$$ 11. Let base $= y + 12$, altitude $= y - 13$, difference $= 25$ (confirm): $$(y + 12) - (y - 13) = 12 + 13 = 25$$ 12. Altitude is: $$y - 13$$ --- 7. **Problem:** Mico's present age $= x$, product of present and two years ago ages is $x^2 - 38x + 360$. Given $x=58$, find current age. 8. Substitute $x=58$: $$58^2 - 38 \times 58 + 360 = 3364 - 2204 + 360 = 1520$$ 9. The product of ages is 1520; Mico's present age is 58. 10. Hence, Mico is 58 years old now. --- 8. **Problem:** Area of rectangular frame is $10y^2 + 33y - 19$ mm$^2$. Length is $3y + 20$ mm longer than width. Find length. 9. Let width = $w$. Then length = $w + 3y + 20$ 10. Area: $$A = w (w + 3y + 20) = 10y^2 + 33y - 19$$ 11. Rearrange: $$w^2 + (3y + 20) w - (10y^2 + 33y - 19) = 0$$ 12. Solve quadratic in $w$: Discriminant $$\Delta = (3y + 20)^2 + 4(10y^2 + 33y - 19)$$ Calculate: $$(3y + 20)^2 = 9y^2 + 120y + 400$$ $$4(10y^2 + 33y - 19) = 40y^2 + 132y - 76$$ Sum: $$49y^2 + 252y + 324$$ 13. Solve for $w$: $$w = \frac{-(3y + 20) \pm \sqrt{49y^2 + 252y + 324}}{2}$$ 14. Length: $$l = w + 3y + 20 = \frac{-(3y + 20) \pm \sqrt{49y^2 + 252y + 324}}{2} + 3y + 20 = \frac{(3y + 20) \pm \sqrt{49y^2 + 252y + 324}}{2}$$ --- 9. **Problem:** Number of trees planted is $d^2 + 6d - 432$. For $d=45$, larger factor is number of trees per row. Find trees per row. 10. Factor: $$d^2 + 6d - 432$$ Find two numbers multiplying to $-432$ and summing to 6: 24 and -18. Factorization: $$(d + 24)(d - 18)$$ 11. Substitute $d=45$: $$(45 + 24) = 69,\quad (45 - 18) = 27$$ 12. Larger factor is 69, so number of trees in a row is 69. --- 10. **Problem:** Find sum of all positive integers $n$ such that $$x^2 + nx^2 + 180 = (x + p)(x + q)$$ where $p, q$ positive integers. 11. Combine like terms: $$x^2 + n x^2 + 180 = (1 + n) x^2 + 180$$ 12. Right side: $$(x + p)(x + q) = x^2 + (p + q) x + p q$$ 13. Since left side has no $x$ term, coefficient of $x$ on right must be zero: $$p + q = 0$$ But $p, q$ positive, so contradiction unless $x$ term is zero. 14. Possibly typo; assume the left is $x^2 + n x +180$: Then: $$(x + p)(x + q) = x^2 + (p + q) x + p q$$ Equate coefficients: $$n = p + q, \quad 180 = p q$$ 15. Find pairs $(p,q)$ of positive integers with $p q = 180$, sum $n = p + q$ Possible factor pairs and sums: (1,180) sum=181 (2,90) sum=92 (3,60) sum=63 (4,45) sum=49 (5,36) sum=41 (6,30) sum=36 (9,20) sum=29 (10,18) sum=28 (12,15) sum=27 16. Sum all such $n$: $$181 + 92 + 63 + 49 + 41 + 36 + 29 + 28 + 27 = 546$$ --- 11. **Problem:** If $x$ is rational, $x > 0$, and $$x = \sqrt{143} + 2\sqrt{143} + 2\sqrt{143} + 2\sqrt{...}$$ Find the minimum value of $x$. 12. Interpretation: An infinite nested radical: $$x = \sqrt{143} + 2x$$ (rearranging the infinite pattern) 13. We can write: $$x = \sqrt{143} + 2x $$ is not exact; but standard form is: $$x = \sqrt{143 + 2x}$$ 14. Square both sides: $$x^2 = 143 + 2x$$ 15. Rearrange quadratic: $$x^2 - 2x - 143 = 0$$ 16. Solve: $$x = \frac{2 \pm \sqrt{4 + 572}}{2} = \frac{2 \pm \sqrt{576}}{2} = \frac{2 \pm 24}{2}$$ 17. Positive solutions: $$x = \frac{2 + 24}{2} = 13,\quad x = \frac{2 - 24}{2} = -11$$ 18. Since $x > 0$, minimum value is $13$. --- 12. **Problem:** Find value of $$\frac{20242024 - 40484049}{(20242023)^2 - (20242024)(20242021)}$$ 13. Numerator: $$20242024 - 40484049 = -20242025$$ 14. Denominator is difference of squares type: $$(20242023)^2 - (20242024)(20242021)$$ Rewrite: $$a = 20242023$$ $$b = 20242024 = a + 1$$ $$c = 20242021 = a - 2$$ So denominator: $$a^2 - (a+1)(a-2) = a^2 - (a^2 - 2a + a - 2) = a^2 - (a^2 - a - 2) = a^2 - a^2 + a + 2 = a + 2$$ 15. Substitute $a = 20242023$: $$20242023 + 2 = 20242025$$ 16. Final value: $$\frac{-20242025}{20242025} = -1$$