Square Completion Aptitude

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### Introduction to Square Completion Square completion is a fundamental algebraic technique used to convert a quadratic expression of the form $ax^2 + bx + c$ into a perfect square trinomial plus a constant, i.e., $a(x-h)^2 + k$. This method is crucial for solving quadratic equations, finding the vertex of parabolas, integrating certain functions, and simplifying expressions in various aptitude tests. **General Formula:** To complete the square for $x^2 + bx$: Add $(\frac{b}{2})^2$ to get $x^2 + bx + (\frac{b}{2})^2 = (x + \frac{b}{2})^2$. If the coefficient of $x^2$ is not 1 (i.e., $ax^2 + bx + c$): 1. Factor out $a$ from the $x^2$ and $x$ terms: $a(x^2 + \frac{b}{a}x) + c$. 2. Complete the square inside the parenthesis for $x^2 + \frac{b}{a}x$: Add $(\frac{b}{2a})^2$ to get $a(x^2 + \frac{b}{a}x + (\frac{b}{2a})^2) - a(\frac{b}{2a})^2 + c$. 3. Simplify: $a(x + \frac{b}{2a})^2 - \frac{b^2}{4a} + c$. ### Solving Quadratic Equations Square completion provides a direct way to solve $ax^2 + bx + c = 0$. #### Steps: 1. Divide by $a$ (if $a \neq 1$): $x^2 + \frac{b}{a}x + \frac{c}{a} = 0$. 2. Move the constant term to the right side: $x^2 + \frac{b}{a}x = -\frac{c}{a}$. 3. Complete the square on the left side by adding $(\frac{b}{2a})^2$ to both sides: $x^2 + \frac{b}{a}x + (\frac{b}{2a})^2 = -\frac{c}{a} + (\frac{b}{2a})^2$. 4. Factor the left side and simplify the right side: $(x + \frac{b}{2a})^2 = \frac{b^2}{4a^2} - \frac{c}{a} = \frac{b^2 - 4ac}{4a^2}$. 5. Take the square root of both sides: $x + \frac{b}{2a} = \pm\sqrt{\frac{b^2 - 4ac}{4a^2}} = \pm\frac{\sqrt{b^2 - 4ac}}{2a}$. 6. Isolate $x$: $x = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ (Quadratic Formula). #### Example: Solve $2x^2 - 8x + 6 = 0$ 1. Divide by 2: $x^2 - 4x + 3 = 0$. 2. Move constant: $x^2 - 4x = -3$. 3. Complete the square: $x^2 - 4x + (-4/2)^2 = -3 + (-4/2)^2$. $x^2 - 4x + 4 = -3 + 4$. 4. Factor: $(x - 2)^2 = 1$. 5. Take square root: $x - 2 = \pm\sqrt{1}$. $x - 2 = \pm 1$. 6. Isolate $x$: $x = 2 \pm 1$. So, $x = 3$ or $x = 1$. ### Finding Vertex of Parabolas For a quadratic function $y = ax^2 + bx + c$, completing the square transforms it into the vertex form $y = a(x-h)^2 + k$, where $(h, k)$ is the vertex of the parabola. #### Steps: 1. Factor out $a$ from the $x^2$ and $x$ terms: $y = a(x^2 + \frac{b}{a}x) + c$. 2. Complete the square inside the parenthesis: $y = a(x^2 + \frac{b}{a}x + (\frac{b}{2a})^2) - a(\frac{b}{2a})^2 + c$. 3. Simplify: $y = a(x + \frac{b}{2a})^2 - \frac{b^2}{4a} + c$. This is $y = a(x - (-\frac{b}{2a}))^2 + (c - \frac{b^2}{4a})$. 4. The vertex is $(h, k) = (-\frac{b}{2a}, c - \frac{b^2}{4a})$ or $(-\frac{b}{2a}, \frac{4ac - b^2}{4a})$. #### Example: Find the vertex of $y = 3x^2 - 6x + 5$ 1. Factor out 3: $y = 3(x^2 - 2x) + 5$. 2. Complete the square: $y = 3(x^2 - 2x + (-2/2)^2) - 3(-2/2)^2 + 5$. $y = 3(x^2 - 2x + 1) - 3(1) + 5$. 3. Simplify: $y = 3(x - 1)^2 - 3 + 5$. $y = 3(x - 1)^2 + 2$. 4. The vertex is $(h, k) = (1, 2)$. ### Minimizing/Maximizing Expressions The vertex form $a(x-h)^2 + k$ is very useful for finding the minimum or maximum value of a quadratic expression. - If $a > 0$, the parabola opens upwards, and the vertex $(h, k)$ represents the minimum point. The minimum value of the expression is $k$. - If $a 0$, the parabola opens upwards. The minimum value occurs at $x = -3$, and the minimum value is $1$. #### Example: Find the maximum value of $g(x) = -2x^2 + 4x - 5$ 1. Factor out -2: $g(x) = -2(x^2 - 2x) - 5$. 2. Complete the square: $g(x) = -2(x^2 - 2x + (-2/2)^2) - (-2)(-2/2)^2 - 5$. $g(x) = -2(x^2 - 2x + 1) - (-2)(1) - 5$. $g(x) = -2(x - 1)^2 + 2 - 5$. 3. Simplify: $g(x) = -2(x - 1)^2 - 3$. 4. Since $a=-2 ### Integration (Calculus context) Square completion is often used to simplify integrands involving quadratic expressions in the denominator or under a square root, allowing the use of standard integration formulas. #### Common Forms: - $\int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \arctan(\frac{x}{a}) + C$ - $\int \frac{1}{\sqrt{a^2 - x^2}} dx = \arcsin(\frac{x}{a}) + C$ - $\int \frac{1}{\sqrt{x^2 \pm a^2}} dx = \ln|x + \sqrt{x^2 \pm a^2}| + C$ #### Example: Evaluate $\int \frac{1}{x^2 + 4x + 8} dx$ 1. Complete the square for the denominator: $x^2 + 4x + 8 = (x^2 + 4x + (4/2)^2) - (4/2)^2 + 8$. $x^2 + 4x + 8 = (x^2 + 4x + 4) - 4 + 8$. $x^2 + 4x + 8 = (x + 2)^2 + 4$. 2. Substitute into the integral: $\int \frac{1}{(x + 2)^2 + 2^2} dx$. 3. Let $u = x + 2$, so $du = dx$. $\int \frac{1}{u^2 + 2^2} du$. 4. Use the formula $\int \frac{1}{u^2 + a^2} du = \frac{1}{a} \arctan(\frac{u}{a}) + C$: $\frac{1}{2} \arctan(\frac{u}{2}) + C$. 5. Substitute back $u = x + 2$: $\frac{1}{2} \arctan(\frac{x + 2}{2}) + C$. ### Geometry: Circles and Ellipses Square completion is essential for transforming general equations of conic sections into their standard forms, making it easy to identify their properties (center, radius, axes). #### Circle Equation: $(x-h)^2 + (y-k)^2 = r^2$ #### Ellipse Equation: $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$ #### Example: Find the center and radius of the circle $x^2 + y^2 - 6x + 10y + 18 = 0$ 1. Group $x$ terms and $y$ terms, and move constant: $(x^2 - 6x) + (y^2 + 10y) = -18$. 2. Complete the square for $x$ terms: $x^2 - 6x + (-6/2)^2 = x^2 - 6x + 9 = (x-3)^2$. 3. Complete the square for $y$ terms: $y^2 + 10y + (10/2)^2 = y^2 + 10y + 25 = (y+5)^2$. 4. Add the added constants to the right side of the equation: $(x^2 - 6x + 9) + (y^2 + 10y + 25) = -18 + 9 + 25$. $(x-3)^2 + (y+5)^2 = 16$. 5. From the standard form, the center is $(h, k) = (3, -5)$ and the radius is $r = \sqrt{16} = 4$. #### Example: Convert $4x^2 + 9y^2 + 16x - 18y - 11 = 0$ to standard ellipse form 1. Group terms and move constant: $(4x^2 + 16x) + (9y^2 - 18y) = 11$. 2. Factor out coefficients: $4(x^2 + 4x) + 9(y^2 - 2y) = 11$. 3. Complete the square for $x$ terms: $4(x^2 + 4x + (4/2)^2) - 4(4/2)^2$. $4(x^2 + 4x + 4) - 4(4) = 4(x+2)^2 - 16$. 4. Complete the square for $y$ terms: $9(y^2 - 2y + (-2/2)^2) - 9(-2/2)^2$. $9(y^2 - 2y + 1) - 9(1) = 9(y-1)^2 - 9$. 5. Substitute back into the equation: $4(x+2)^2 - 16 + 9(y-1)^2 - 9 = 11$. $4(x+2)^2 + 9(y-1)^2 = 11 + 16 + 9$. $4(x+2)^2 + 9(y-1)^2 = 36$. 6. Divide by 36 to get 1 on the right side: $\frac{4(x+2)^2}{36} + \frac{9(y-1)^2}{36} = \frac{36}{36}$. $\frac{(x+2)^2}{9} + \frac{(y-1)^2}{4} = 1$. This is the standard form of an ellipse centered at $(-2, 1)$ with semi-major axis $a=3$ and semi-minor axis $b=2$. ### Advanced Aptitude Problems Square completion can be used in more complex scenarios, such as finding minimum/maximum values of multi-variable expressions or simplifying expressions under square roots. #### Example: Find the minimum value of $E = x^2 + y^2 - 4x + 6y + 15$ 1. Group $x$ terms and $y$ terms: $E = (x^2 - 4x) + (y^2 + 6y) + 15$. 2. Complete the square for $x$ terms: $(x^2 - 4x + (-4/2)^2) - (-4/2)^2 = (x-2)^2 - 4$. 3. Complete the square for $y$ terms: $(y^2 + 6y + (6/2)^2) - (6/2)^2 = (y+3)^2 - 9$. 4. Substitute back into the expression: $E = (x-2)^2 - 4 + (y+3)^2 - 9 + 15$. 5. Simplify: $E = (x-2)^2 + (y+3)^2 + 2$. 6. Since $(x-2)^2 \ge 0$ and $(y+3)^2 \ge 0$, the minimum value of $E$ occurs when $(x-2)^2 = 0$ (i.e., $x=2$) and $(y+3)^2 = 0$ (i.e., $y=-3$). The minimum value is $0 + 0 + 2 = 2$. #### Example: Simplify $\sqrt{x^2 - 10x + 29}$ 1. Complete the square for the expression inside the square root: $x^2 - 10x + 29 = (x^2 - 10x + (-10/2)^2) - (-10/2)^2 + 29$. $x^2 - 10x + 29 = (x^2 - 10x + 25) - 25 + 29$. $x^2 - 10x + 29 = (x-5)^2 + 4$. 2. Substitute back into the square root: $\sqrt{(x-5)^2 + 4}$. This simplified form is often useful in calculus (e.g., trigonometric substitution) or geometry.