Solving Quadratic Equations By Completing The Square Method

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This article delves into the method of completing the square, a powerful technique for solving quadratic equations. We will explore the step-by-step process with detailed explanations and examples to ensure a comprehensive understanding. Completing the square is not only a method for finding solutions but also a foundational concept for deriving the quadratic formula and understanding the vertex form of a parabola. Mastering this technique provides a robust tool for tackling various algebraic problems.

Understanding Quadratic Equations

Before diving into the completing the square method, it's crucial to understand the basics of quadratic equations. A quadratic equation is a polynomial equation of the second degree. The general form of a quadratic equation is:

ax2+bx+c=0ax^2 + bx + c = 0

Where a, b, and c are constants, and x represents the variable. The coefficient a cannot be zero, as this would make the equation linear rather than quadratic. The solutions to a quadratic equation, also known as roots or zeros, are the values of x that satisfy the equation. These solutions represent the points where the parabola defined by the quadratic equation intersects the x-axis.

Quadratic equations can have two distinct real solutions, one repeated real solution, or two complex solutions. The nature of the solutions depends on the discriminant, which is given by the formula:

D=b2−4acD = b^2 - 4ac

  • If D > 0, the equation has two distinct real solutions.
  • If D = 0, the equation has one repeated real solution.
  • If D < 0, the equation has two complex solutions.

There are several methods to solve quadratic equations, including factoring, using the quadratic formula, and completing the square. Each method has its strengths and is suitable for different types of quadratic equations. In this article, we will focus on the completing the square method.

The Method of Completing the Square

Completing the square is a technique used to convert a quadratic equation from its standard form ($ax^2 + bx + c = 0$) into the vertex form ($a(x - h)^2 + k = 0$), which makes it easier to solve for x. The vertex form reveals the vertex of the parabola represented by the quadratic equation, which is the point (h, k). This method is particularly useful when the quadratic equation cannot be easily factored.

The steps involved in completing the square are as follows:

  1. Divide by a: If a is not equal to 1, divide the entire equation by a. This step ensures that the coefficient of the $x^2$ term is 1, which is necessary for completing the square.
  2. Move the constant term: Move the constant term (c) to the right side of the equation. This isolates the terms with x on the left side.
  3. Complete the square: Take half of the coefficient of the x term (which is b), square it, and add it to both sides of the equation. This step creates a perfect square trinomial on the left side.
  4. Factor the perfect square trinomial: Factor the left side of the equation as a perfect square, which will be in the form $(x + rac{b}{2})^2$.
  5. Solve for x: Take the square root of both sides of the equation. Remember to consider both positive and negative square roots. Isolate x to find the solutions.

Let's illustrate this method with the examples provided.

Example 1: Solving $k^2 + 7k + 12 = 0$

Step 1: Verify the coefficient of $k^2$

In this equation, the coefficient of $k^2$ is 1, so we can proceed to the next step. If the coefficient were not 1, we would need to divide the entire equation by that coefficient.

Step 2: Move the Constant Term

Subtract 12 from both sides of the equation to move the constant term to the right side:

k2+7k=−12k^2 + 7k = -12

Step 3: Complete the Square

To complete the square, we need to add a value to both sides of the equation that will make the left side a perfect square trinomial. This value is determined by taking half of the coefficient of the k term (which is 7), squaring it, and adding the result to both sides.

Half of 7 is $ rac{7}{2}$, and squaring it gives us $( rac{7}{2})^2 = rac{49}{4}$. Add this to both sides:

k^2 + 7k + rac{49}{4} = -12 + rac{49}{4}

Step 4: Factor the Perfect Square Trinomial

The left side of the equation is now a perfect square trinomial, which can be factored as:

(k + rac{7}{2})^2

On the right side, we need to combine the terms:

-12 + rac{49}{4} = - rac{48}{4} + rac{49}{4} = rac{1}{4}

So the equation becomes:

(k + rac{7}{2})^2 = rac{1}{4}

Step 5: Solve for k

Take the square root of both sides of the equation:

k + rac{7}{2} = \\\pm rac{1}{2}

Subtract $ rac{7}{2}$ from both sides to isolate k:

k = - rac{7}{2} \\\pm rac{1}{2}

This gives us two possible solutions:

k_1 = - rac{7}{2} + rac{1}{2} = - rac{6}{2} = -3

k_2 = - rac{7}{2} - rac{1}{2} = - rac{8}{2} = -4

Therefore, the solutions to the quadratic equation $k^2 + 7k + 12 = 0$ are k = -3 and k = -4.

Example 2: Solving $y^2 + 6y - 27 = 0$

Step 1: Verify the coefficient of $y^2$

The coefficient of $y^2$ is 1, so we can proceed to the next step.

Step 2: Move the Constant Term

Add 27 to both sides of the equation:

y2+6y=27y^2 + 6y = 27

Step 3: Complete the Square

Take half of the coefficient of the y term (which is 6), square it, and add it to both sides. Half of 6 is 3, and squaring it gives us 9. Add 9 to both sides:

y2+6y+9=27+9y^2 + 6y + 9 = 27 + 9

Step 4: Factor the Perfect Square Trinomial

The left side of the equation is a perfect square trinomial, which can be factored as:

(y+3)2(y + 3)^2

On the right side, we have:

27+9=3627 + 9 = 36

So the equation becomes:

(y+3)2=36(y + 3)^2 = 36

Step 5: Solve for y

Take the square root of both sides of the equation:

y+3=±6y + 3 = \\\pm6

Subtract 3 from both sides to isolate y:

y=−3±6y = -3 \\\pm 6

This gives us two possible solutions:

y1=−3+6=3y_1 = -3 + 6 = 3

y2=−3−6=−9y_2 = -3 - 6 = -9

Therefore, the solutions to the quadratic equation $y^2 + 6y - 27 = 0$ are y = 3 and y = -9.

Example 3: Solving $9z^2 - 9z - 12 = 0$

Step 1: Divide by a

In this case, the coefficient of $z^2$ is 9, which is not 1. Therefore, we need to divide the entire equation by 9:

z^2 - z - rac{4}{3} = 0

Step 2: Move the Constant Term

Add $ rac{4}{3}$ to both sides of the equation:

z^2 - z = rac{4}{3}

Step 3: Complete the Square

Take half of the coefficient of the z term (which is -1), square it, and add it to both sides. Half of -1 is $- rac{1}{2}$, and squaring it gives us $( rac{-1}{2})^2 = rac{1}{4}$. Add $ rac{1}{4}$ to both sides:

z^2 - z + rac{1}{4} = rac{4}{3} + rac{1}{4}

Step 4: Factor the Perfect Square Trinomial

The left side of the equation is a perfect square trinomial, which can be factored as:

(z - rac{1}{2})^2

On the right side, we need to combine the terms:

rac{4}{3} + rac{1}{4} = rac{16}{12} + rac{3}{12} = rac{19}{12}

So the equation becomes:

(z - rac{1}{2})^2 = rac{19}{12}

Step 5: Solve for z

Take the square root of both sides of the equation:

z - rac{1}{2} = \\\pm rac{\sqrt{19}}{\sqrt{12}} = \\\pm rac{\sqrt{19}}{2\sqrt{3}}

Rationalize the denominator by multiplying the numerator and denominator by $\\sqrt{3}$:

z - rac{1}{2} = \\\pm rac{\sqrt{57}}{6}

Add $ rac{1}{2}$ to both sides to isolate z:

z = rac{1}{2} \\\pm rac{\sqrt{57}}{6}

This gives us two possible solutions:

z_1 = rac{1}{2} + rac{\sqrt{57}}{6} = rac{3 + \\sqrt{57}}{6}

z_2 = rac{1}{2} - rac{\sqrt{57}}{6} = rac{3 - \\sqrt{57}}{6}

Therefore, the solutions to the quadratic equation $9z^2 - 9z - 12 = 0$ are $z = rac{3 + \sqrt{57}}{6}$ and $z = rac{3 - \sqrt{57}}{6}$. Note that we have replaced the original variable '9' with 'z' to adhere to standard algebraic notation.

Example 4: Solving $u^2 + u - 90 = 0$

Step 1: Verify the coefficient of $u^2$

The coefficient of $u^2$ is 1, so we can proceed to the next step.

Step 2: Move the Constant Term

Add 90 to both sides of the equation:

u2+u=90u^2 + u = 90

Step 3: Complete the Square

Take half of the coefficient of the u term (which is 1), square it, and add it to both sides. Half of 1 is $ rac{1}{2}$, and squaring it gives us $( rac{1}{2})^2 = rac{1}{4}$. Add $ rac{1}{4}$ to both sides:

u^2 + u + rac{1}{4} = 90 + rac{1}{4}

Step 4: Factor the Perfect Square Trinomial

The left side of the equation is a perfect square trinomial, which can be factored as:

(u + rac{1}{2})^2

On the right side, we need to combine the terms:

90 + rac{1}{4} = rac{360}{4} + rac{1}{4} = rac{361}{4}

So the equation becomes:

(u + rac{1}{2})^2 = rac{361}{4}

Step 5: Solve for u

Take the square root of both sides of the equation:

u + rac{1}{2} = \\\pm rac{\sqrt{361}}{\sqrt{4}} = \\\pm rac{19}{2}

Subtract $ rac{1}{2}$ from both sides to isolate u:

u = - rac{1}{2} \\\pm rac{19}{2}

This gives us two possible solutions:

u_1 = - rac{1}{2} + rac{19}{2} = rac{18}{2} = 9

u_2 = - rac{1}{2} - rac{19}{2} = - rac{20}{2} = -10

Therefore, the solutions to the quadratic equation $u^2 + u - 90 = 0$ are u = 9 and u = -10.

Conclusion

In this article, we have explored the method of completing the square for solving quadratic equations. This technique involves transforming the quadratic equation into vertex form, which allows us to easily solve for the variable. We have demonstrated the step-by-step process with several examples, covering cases where the leading coefficient is 1 and cases where it is not. Completing the square is a valuable tool in algebra and provides a deeper understanding of quadratic equations and their solutions. By mastering this method, you can confidently tackle a wide range of quadratic equation problems.