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3 To understand and solve quadratic equations, we can use the method of completing the square. This method converts a quadratic expression into vertex form. For instance, this technique transforms a quadratic expression of the form ax² + bx + c into the vertex form, which is a(x - h)² + k, where (h, k) represents the vertex of the parabola. Hence, the left-hand side becomes a perfect square trinomial, helping in rewriting it in vertex form. In this article, we will explore completing the square in detail.

4 <h2>What is Completing the Square?</h2>

5 What Is Algebra? 🧮 | Simple Explanation with 🎯 Cool Examples for Kids | ✨BrightCHAMPS Math

6 ▶

7 In<a>algebra</a>, completing the<a>square</a>is a technique used to rewrite a<a>quadratic expression</a>in a form that is a<a>perfect square</a>. For example, a quadratic<a>equation</a>like:

8 \(ax^2 + bx + c = 0\)

9 It can be rewritten (by completing the square) in the form:

10 \(a(x + p)^2 + q = 0\)

11 Where p and q are<a>numbers</a>we calculate during the process.

12 <h2>How to Complete the Square Method?</h2>

13 We can use the process of completing the square to find the roots or zeros of a quadratic equation or<a>quadratic polynomial</a>, and to factorize the equation. When the<a>expression</a>is not possible to factorize, this technique plays an important role.

14 For instance, \(x^2 + 2x + 3\) cannot be factorized using<a>real numbers</a>, because there are no two numbers that add to 2 and multiply to 3. In such cases, we use completing the square to rewrite it in a new form:

15 \(a (x + m)^2 + n \)

16 This form helps us express the quadratic as a perfect square plus a<a>constant</a>.

17 Here, by rewriting the expression as \((x + m)\), we complete the square.

18 <h2>What are the Steps for Completing the Square?</h2>

19 We have to follow certain steps for the method of completing the square.

20 Step 1:Express the quadratic equation as \(x^2 + bx + c\). Here, 1 must be the<a>coefficient</a>of \(x^2\). If it is not 1, the number will be a<a>common factor</a>and placed outside.

21 Step 2:Find the half of the coefficient of \(x\).

22 Step 3:Find the square of the number (half of the coefficient).

23 Step 4:Add and subtract the square within the expression to maintain equality.

24 Step 5:To complete the square, factorize the<a>polynomial</a>and use the<a>algebraic identity</a>.

25 \(x^2 + 2xy + y^2 = (x + y)^2\)

26 Or

27 \(x^2 - 2xy + y^2 = (x - y)^2\)

28 Now, let us take an example for better understanding.

29 Complete the square for \(-4x^2 - 8x - 12\).

30 Step 1:First, we need to find if the coefficient is 1. Here, the coefficient of x2 is not 1, so the number (-4) is placed outside as a common factor.

31 \(-4x^2 - 8x - 12 = -4 (x^2 + 2x + 3)\)

32 However, the coefficient of x2 is now 1.

33 Step 2:Next, we need to find half of the coefficient of x. Coefficient of x = 2 Half of 2 = 1

34 Step 3:Find the square of 1. \(1^2 = 1\)

35 Step 4:Add and subtract the square to the x2<a>term</a>.

36 \(-4 (x^2 + 2x + 3) = -4 (x^2 + 2x + 1 - 1 + 3) \)

37 Step 5:Factorize the polynomial and apply the algebraic identity.

38 \(x^2 + 2xy + y^2 = (x + y)^2\)

39 Here, the first three terms: \(x^2 + 2x + 1 \)

40 Last two terms: \(-1 + 3 \)

41 \(x^2 + 2x + 1 = (x + 1)^2 \)

42 \(-4 (x^2 + 2x + 1 - 1 + 3) = -4 ((x + 1)² - 1 + 3)\)

43 Step 6:Next, simplify the last two terms: \(-1 + 3 \) \(-1 + 3 = 2 \)

44 Now, the expression becomes:

45 \(-4 (( x + 1)^2 + 2) \)

46 Step 7:Next, we can distribute the -4:

47 \(-4 (x + 1)^2 - 8 \)

48 Hence, the final result is:

49 \(-4x^2 - 8x - 12 = -4 (x + 1)^2 - 8\)

50 Now, this is the completed square form:

51 \(a (x + m)^2 + n \)

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54 <h2>What is the Formula for Completing the Square?</h2>

53 <h2>What is the Formula for Completing the Square?</h2>

55 Converting a quadratic equation or polynomial into a perfect square with an extra constant is known as the process of completing the square.

54 Converting a quadratic equation or polynomial into a perfect square with an extra constant is known as the process of completing the square.

56 For a quadratic expression: \( ax^2 + bx + c = 0 \), where, a, b, and c are real numbers, but a is<a>not equal</a>to 0.

55 For a quadratic expression: \( ax^2 + bx + c = 0 \), where, a, b, and c are real numbers, but a is<a>not equal</a>to 0.

57 The<a>formula</a>for completing the square is:

56 The<a>formula</a>for completing the square is:

58 \(ax² + bx + c = a (x + m)² + n \)

57 \(ax² + bx + c = a (x + m)² + n \)

59 Here,

58 Here,

60 \(m = \frac{b}{2a} \)

59 \(m = \frac{b}{2a} \)

61 \(n = c - (\frac{b^2}{4a}) \)

60 \(n = c - (\frac{b^2}{4a}) \)

62 When completing the square of a given expression, we first need to find the values of m and n, then substitute these values into the formula.

61 When completing the square of a given expression, we first need to find the values of m and n, then substitute these values into the formula.

63 Now, let us understand the application of the formula for completing the square.

62 Now, let us understand the application of the formula for completing the square.

64 For example, \(x^2 - 4x - 8 = 0 \)

63 For example, \(x^2 - 4x - 8 = 0 \)

65 Step 1:Now, we can apply the formula for completing the square:

64 Step 1:Now, we can apply the formula for completing the square:

66 \(ax² + bx + c = a (x + m)² + n \)

65 \(ax² + bx + c = a (x + m)² + n \)

67 Step 2:Here,

66 Step 2:Here,

68 a = 1 b = -4 c = -8

67 a = 1 b = -4 c = -8

69 Step 3:Now, we can find the value of m and n.

68 Step 3:Now, we can find the value of m and n.

70 \(m = \frac{b}{2a} = \frac{-4}{2} (1) = -2\)

69 \(m = \frac{b}{2a} = \frac{-4}{2} (1) = -2\)

71 \(n = c - (\frac{b^2}{4a}) = -8 - (\frac{(-4)^2}{4} (1))\)

70 \(n = c - (\frac{b^2}{4a}) = -8 - (\frac{(-4)^2}{4} (1))\)

72 \(n = -8 - (\frac{16}{4}) = -8 - 4 \)

71 \(n = -8 - (\frac{16}{4}) = -8 - 4 \)

73 \(n = -12 \)

72 \(n = -12 \)

74 Step 4:So, the expression becomes:

73 Step 4:So, the expression becomes:

75 \(x^2 - 4x - 8 = (x - 2)^2 - 12 \)

74 \(x^2 - 4x - 8 = (x - 2)^2 - 12 \)

76 Step 5:Now, we can solve the expression.

75 Step 5:Now, we can solve the expression.

77 \((x - 2)^2 - 12 = 0\)

76 \((x - 2)^2 - 12 = 0\)

78 \((x - 2)^2 = 12 \)

77 \((x - 2)^2 = 12 \)

79 \(x - 2 = √12 = 2√3 \)

78 \(x - 2 = √12 = 2√3 \)

80 \(x = 2 ∓ 2√3 \)

79 \(x = 2 ∓ 2√3 \)

81 Therefore, \(x = 2 + 2√3\) or \(x = 2 - 2√3\)

80 Therefore, \(x = 2 + 2√3\) or \(x = 2 - 2√3\)

82 <ul><li>For geometric visualization, divide the rectangle. So, b / 2a will be the length of each rectangle. </li>

81 <ul><li>For geometric visualization, divide the rectangle. So, b / 2a will be the length of each rectangle. </li>

83 </ul><ul><li>Fix one rectangle to the right side of the square and the next one to the bottom of the square. </li>

82 </ul><ul><li>Fix one rectangle to the right side of the square and the next one to the bottom of the square. </li>

84 <li>To complete the geometric square, we need to add a square of area [ (b / 2a)2 ] to x2 + (b / a) x. To retain the value of the expression, we must subtract it. Thus, to complete the square:\(x^2 + (\frac{b}{a}) x = x^2 + (\frac{b}{a}) x +(\frac{b}{2a})^2 - (\frac{b}{2a})^2\)

83 <li>To complete the geometric square, we need to add a square of area [ (b / 2a)2 ] to x2 + (b / a) x. To retain the value of the expression, we must subtract it. Thus, to complete the square:\(x^2 + (\frac{b}{a}) x = x^2 + (\frac{b}{a}) x +(\frac{b}{2a})^2 - (\frac{b}{2a})^2\)

85 \(x^2 + (\frac{b}{a}) x = x^2 + (\frac{b}{a}) x + (\frac{b}{2a})2 - \frac{b^2}{4a^2}\)

84 \(x^2 + (\frac{b}{a}) x = x^2 + (\frac{b}{a}) x + (\frac{b}{2a})2 - \frac{b^2}{4a^2}\)

86 </li>

85 </li>

87 </ul><ul><li>Multiplying and dividing \((\frac{b}{a}) x\) with 2 gives, \(x^2 + (2x × \frac{b}{2a}) + (\frac{b}{2a})^2 - \frac{b^2}{4a^2}\)

86 </ul><ul><li>Multiplying and dividing \((\frac{b}{a}) x\) with 2 gives, \(x^2 + (2x × \frac{b}{2a}) + (\frac{b}{2a})^2 - \frac{b^2}{4a^2}\)

88 </li>

87 </li>

89 <li>Now, we can use the identity,\(x^2 + 2xy + y^2 = (x + y)^2 \)

88 <li>Now, we can use the identity,\(x^2 + 2xy + y^2 = (x + y)^2 \)

90 The equation above can be expressed as,

89 The equation above can be expressed as,

91 \( x^2 + (\frac{b}{a}) x = (x + \frac{b}{2a})^2 - (\frac{b^2}{4a^2}) \)

90 \( x^2 + (\frac{b}{a}) x = (x + \frac{b}{2a})^2 - (\frac{b^2}{4a^2}) \)

92 </li>

91 </li>

93 </ul>Now, we had:

92 </ul>Now, we had:

94 \(ax^2 + bx + c = a (x^2 + \frac{b}{ax}) + c\)

93 \(ax^2 + bx + c = a (x^2 + \frac{b}{ax}) + c\)

95 <ul><li>Next, substitute the completed square form: \(a ((x +\frac{b}{2a})^2 - b^2 / 4a^2) + c \)

94 <ul><li>Next, substitute the completed square form: \(a ((x +\frac{b}{2a})^2 - b^2 / 4a^2) + c \)

96 </li>

95 </li>

97 </ul><ul><li>Now, distribute the ‘a’: \(a (x + \frac{b}{2a})^2 - a (\frac{b^2}{4a^2}) + c \)

96 </ul><ul><li>Now, distribute the ‘a’: \(a (x + \frac{b}{2a})^2 - a (\frac{b^2}{4a^2}) + c \)

98 </li>

97 </li>

99 <li>Then, simplify:\(a (\frac{b^2}{4a^2}) = \frac{b^2}{4a} \)

98 <li>Then, simplify:\(a (\frac{b^2}{4a^2}) = \frac{b^2}{4a} \)

100 </li>

99 </li>

101 </ul><ul><li>Therefore, the expression becomes:\(a (x +\frac{ b}{2a})^2 - \frac{b^2}{4a} + c\)

100 </ul><ul><li>Therefore, the expression becomes:\(a (x +\frac{ b}{2a})^2 - \frac{b^2}{4a} + c\)

102 </li>

101 </li>

103 <li>Next, arrange the constants: \(a (x + \frac{b}{2a})^2 - (c - \frac{b^2}{4a})\)

102 <li>Next, arrange the constants: \(a (x + \frac{b}{2a})^2 - (c - \frac{b^2}{4a})\)

104 Thus, \(ax^2 + bx + c = a (x + \frac{b}{2a})^2 - (c - \frac{b^2}{4a})\)

103 Thus, \(ax^2 + bx + c = a (x + \frac{b}{2a})^2 - (c - \frac{b^2}{4a})\)

105 </li>

104 </li>

106 </ul><ul><li>The expression follows the form: \(a (x + m)^2 + n \)

105 </ul><ul><li>The expression follows the form: \(a (x + m)^2 + n \)

107 Where, \(m = \frac{b}{2a}\) and \(n = c - \frac{b^2}{4a}\)

106 Where, \(m = \frac{b}{2a}\) and \(n = c - \frac{b^2}{4a}\)

108 </li>

107 </li>

109 </ul><h2>Derivation of Completing the Square Formula</h2>

108 </ul><h2>Derivation of Completing the Square Formula</h2>

110 Step 1:Start with a general quadratic equation

109 Step 1:Start with a general quadratic equation

111 \( ax² + bx + c = 0,\) where a ≠ 0.

110 \( ax² + bx + c = 0,\) where a ≠ 0.

112 Step 2: Divide the entire equation by a to make the coefficient of x² equal to 1:

111 Step 2: Divide the entire equation by a to make the coefficient of x² equal to 1:

113 \( x^2 + \frac{b}{a}x + \frac{c}{a} = 0\)

112 \( x^2 + \frac{b}{a}x + \frac{c}{a} = 0\)

114 Step 3: Move the constant term to the right side:

113 Step 3: Move the constant term to the right side:

115 \( x^2 + \frac{b}{a}x = -\frac{c}{a} \)

114 \( x^2 + \frac{b}{a}x = -\frac{c}{a} \)

116 Step 4: Take half of the coefficient of x, square it, and add it to both sides.

115 Step 4: Take half of the coefficient of x, square it, and add it to both sides.

117 Half of (b/a) is b/(2a).

116 Half of (b/a) is b/(2a).

118 Square it: (b/2a)² = b²/(4a²).

117 Square it: (b/2a)² = b²/(4a²).

119 Add to both sides:

118 Add to both sides:

120 \( x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} = -\frac{c}{a} + \frac{b^2}{4a^2} \)

119 \( x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} = -\frac{c}{a} + \frac{b^2}{4a^2} \)

121 Step 5:Write the left side as a perfect square:

120 Step 5:Write the left side as a perfect square:

122 \( \left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2} \)

121 \( \left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2} \)

123 Step 6: Take square roots on both sides:

122 Step 6: Take square roots on both sides:

124 \( x + \frac{b}{2a} = \pm \frac{\sqrt{b^2 - 4ac}}{2a} \)

123 \( x + \frac{b}{2a} = \pm \frac{\sqrt{b^2 - 4ac}}{2a} \)

125 Step 7: Solve for x:

124 Step 7: Solve for x:

126 \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

125 \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

127 This gives the quadratic formula.

126 This gives the quadratic formula.

128 <h2>Tips and Tricks to Master Completing the Squares</h2>

127 <h2>Tips and Tricks to Master Completing the Squares</h2>

129 Completing the square is a fundamental algebraic technique used to solve<a>quadratic equations</a>, derive the quadratic formula, and analyze parabolas. Here are some useful tips and tricks to master the concept:

128 Completing the square is a fundamental algebraic technique used to solve<a>quadratic equations</a>, derive the quadratic formula, and analyze parabolas. Here are some useful tips and tricks to master the concept:

130 <ul><li>Ensure the coefficient of \(x^2\) is 1. If it's not,<a>factor</a>it out from the entire equation. This simplifies the process and aligns with standard methods.</li>

129 <ul><li>Ensure the coefficient of \(x^2\) is 1. If it's not,<a>factor</a>it out from the entire equation. This simplifies the process and aligns with standard methods.</li>

131 <li>Divide the linear coefficient by 2 and square it. Take half of the coefficient of 𝑥, square it, and add this value to both sides of the equation. This step creates a perfect square<a>trinomial</a>on the left-hand side. </li>

130 <li>Divide the linear coefficient by 2 and square it. Take half of the coefficient of 𝑥, square it, and add this value to both sides of the equation. This step creates a perfect square<a>trinomial</a>on the left-hand side. </li>

132 <li>They should also remind students to keep the equation balanced by adding and subtracting the same value on both sides. </li>

131 <li>They should also remind students to keep the equation balanced by adding and subtracting the same value on both sides. </li>

133 <li>Take the<a>square root</a>of both sides, remembering to consider both the positive and negative roots. Then, isolate 𝑥 to find the solutions to the quadratic equation. </li>

132 <li>Take the<a>square root</a>of both sides, remembering to consider both the positive and negative roots. Then, isolate 𝑥 to find the solutions to the quadratic equation. </li>

134 <li>Parents and teachers should help students practice different quadratic equations, even with tricky coefficients or complex solutions, to build confidence in completing the square.</li>

133 <li>Parents and teachers should help students practice different quadratic equations, even with tricky coefficients or complex solutions, to build confidence in completing the square.</li>

135 </ul><h2>Common Mistakes and How to Avoid Them on Completing the Square</h2>

134 </ul><h2>Common Mistakes and How to Avoid Them on Completing the Square</h2>

136 Sometimes, completing the square seems tricky for students when they solve quadratic equations. Thus, they often make some errors that lead them to incorrect answers. Here are some common mistakes and their helpful solutions to prevent them.

135 Sometimes, completing the square seems tricky for students when they solve quadratic equations. Thus, they often make some errors that lead them to incorrect answers. Here are some common mistakes and their helpful solutions to prevent them.

137 <h2>Real-Life Applications of Completing the Square</h2>

136 <h2>Real-Life Applications of Completing the Square</h2>

138 Understanding the significance of the method of completing the square is useful since it can be applied in various situations. Here are some real-world applications of the concept.

137 Understanding the significance of the method of completing the square is useful since it can be applied in various situations. Here are some real-world applications of the concept.

139 <ul><li>In engineering and construction, engineers use the method of completing the square to ensure proper stability and balance. For example, if they need to construct a building that has different shapes, such as squares and rectangles, professionals can use the technique to find the dimensions. </li>

138 <ul><li>In engineering and construction, engineers use the method of completing the square to ensure proper stability and balance. For example, if they need to construct a building that has different shapes, such as squares and rectangles, professionals can use the technique to find the dimensions. </li>

140 <li>Astronomers and space scientists can use the technique of completing the square to analyze the satellites and orbits of planets. For instance, to analyze and predict the path of satellites, researchers use the completing the square method. </li>

139 <li>Astronomers and space scientists can use the technique of completing the square to analyze the satellites and orbits of planets. For instance, to analyze and predict the path of satellites, researchers use the completing the square method. </li>

141 <li>Companies and organizations can analyze their<a>profit</a>and break-even points using completing the square process. For example, if the profit is \(P(x) = -4x^2 + 24x - 10\), completing squares can determine the number of items needed to maximize profits.</li>

140 <li>Companies and organizations can analyze their<a>profit</a>and break-even points using completing the square process. For example, if the profit is \(P(x) = -4x^2 + 24x - 10\), completing squares can determine the number of items needed to maximize profits.</li>

142 <li>In electronics, quadratic equations appear in signal amplitude and voltage optimization. Completing the square helps determine peak values and thresholds.</li>

141 <li>In electronics, quadratic equations appear in signal amplitude and voltage optimization. Completing the square helps determine peak values and thresholds.</li>

143 <li>In computer graphics, parabolic curves and quadratic forms are often converted using completing the square to simplify rendering, animation trajectories, or collision detection in games.</li>

142 <li>In computer graphics, parabolic curves and quadratic forms are often converted using completing the square to simplify rendering, animation trajectories, or collision detection in games.</li>

144 - </ul><h3>Problem 1</h3>

143 + </ul><h2>Download Worksheets</h2>

144 + <h3>Problem 1</h3>

145 Complete the square of: x^2 + 8x

146 Okay, lets begin

147 \((x + 4)^2 - 16 \)

148 <h3>Explanation</h3>

149 Here, the given expression is \(x^2 + 8x\)

150 <ul><li>First, we need to find the half of the coefficient of x: \( 8 / 2 = 4 \)

151 </li>

152 <li>Then, square it: 42 = 16 </li>

153 <li>Now, add and subtract 16: \(x^2 + 8x + 16 - 16 \) </li>

154 <li>Here, we take the first three terms: \(x^2 + 8x + 16\) </li>

155 <li>We want to express it in the form: \((x + a)^2 \) \((x + a)^2 = x^2 + 2ax + a^2 \) </li>

156 <li>In the given expression, \(x^2 + 8x + 16\) Coefficient of x is 8. </li>

157 <li>That corresponds to 2a. 2a = 8 a = 4 </li>

158 <li>Therefore, a2 = 42 = 16 </li>

159 <li>So, \(x^2 + 8x + 16 = (x + 4)^2\) </li>

160 <li>Hence, the expression becomes: \( (x + 4)^2 - 16 \)</li>

161 </ul>Thus,\( x^2 + 8x = (x + 4)^2 - 16\)

162 Well explained 👍

163 <h3>Problem 2</h3>

164 Solve the equation, x^2 - 10x

165 Okay, lets begin

166 \((x - 5)^2 - 25\)

167 <h3>Explanation</h3>

168 The given expression is \(x^2 - 10x\)

169 <ul><li>Half of -10 = -5 </li>

170 <li>Square of -5: \((-5)^2 = 25\) </li>

171 <li>Next, add and subtract 25:\(x^2 - 10x + 25 - 25 \)

172 </li>

173 <li>Rewrite it as:\((x - 5)^2 - 25 = (x - 5)^2 \)

174 </li>

175 <li>Hence, the expression becomes:\((x - 5)^2 - 25 \)

176 </li>

177 </ul>Thus, \(x^2 - 10x = (x - 5)^2 - 25\)

178 Well explained 👍

179 <h3>Problem 3</h3>

180 Solve x^2 + 7x + 5 = 0

181 Okay, lets begin

182 \(\frac{(-7 ± √29)}{ 2} \)

183 <h3>Explanation</h3>

184 The given expression is: \(x^2 + 7x + 5\)

185 <ul><li>Move the constant to the right side.: \(x^2 + 7x = - 5\) </li>

186 <li>Take the half of the coefficient of x: Coefficient of x is 7. Half of 7 is: \(\frac{7}{2}\) </li>

187 <li>Next, square it:\( (\frac{7}{2})^2 = \frac{49}{4} \)

188 </li>

189 <li>So, we need to add 49 / 4 to both sides of the equation.\( x^2 + 7x + \frac{49}{4} = -5 + \frac{49}{4} \)

190 </li>

191 <li>Convert -5 to a fraction with denominator 4: \( -5 = \frac{-20}{4} \) </li>

192 <li>Now, add: \(\frac{-20} {4} + \frac{49}{4} = \frac{29}{4}\) </li>

193 <li>Hence, the expression is: \( x^2 + 7x + \frac{49}{4} = \frac{29}{4}\) </li>

194 <li>Here, the left-hand side is a perfect square trinomial:\(x^2 + 7x + \frac{49}{4} = (x + \frac{7}{2})^2 \)

195 </li>

196 <li>Rewrite the equation: \((x + \frac{7}{2})^2 = \frac{29}{4} \)

197 </li>

198 <li>Now, take the square root of both sides:\(x + \frac{7}{2} = √\frac{29} {4} \)

199 \(x + \frac{7}{2} = \frac{√29} {2}\)

200 </li>

201 <li>Next, subtract \(\frac{7}{2}\) from both sides: \(x = - \frac{7}{2} ± \frac{√29}{2} \)

202 </li>

203 </ul>Since the denominators are the same, we can write it as:

204 \(x = \frac{(-7 ± √29)}{2}\)

205 Well explained 👍

206 <h3>Problem 4</h3>

207 Complete the square of the expression: 2x^2 + 12 x

208 Okay, lets begin

209 \(2 (x + 3)^2 - 18 \)

210 <h3>Explanation</h3>

211 <ul><li> First, we need to factor out 2. \(2 (x^2 + 6x) \) </li>

212 <li>Half of 6 = 3 </li>

213 <li>Square it: 32 = 9 </li>

214 <li>Add and subtract inside: \(2 (x^2 + 6x + 9 - 9) \) </li>

215 <li>Rewrite it as: \(2 ((x + 3)^2 - 9) = 2 (x + 3)^2 - 18 \)</li>

216 </ul>Thus, \(2x^2 + 12x = 2 (x + 3)^2 - 18 \)

217 Well explained 👍

218 <h3>Problem 5</h3>

219 Complete the square of the expression: x^2 - 5x

220 Okay, lets begin

221 \((\frac{x - 5}{2})^2 - \frac{25}{4} \)

222 <h3>Explanation</h3>

223 The given expression is:

224 \(x^2 - 5x \)

225 <ul><li>Half of \(-5 = \frac{-5}{2}\) </li>

226 <li>Square it: \( (\frac{-5}{2})^2 = (\frac{25}{4}) \)

227 </li>

228 </ul><ul><li>We add and subtract the square inside the expression, so the value does not change: \(x^2 - 5x = x^2 - 5x + \frac{25}{4} - (\frac{25}{4})\)

229 </li>

230 <li>Now, we apply the identity: \(x^2 - 2xy + y^2 = (x - y)^2 \)

231 </li>

232 </ul><ul><li>Here, \(x^2 - 5x + (\frac{25}{4} )\)is a perfect square trinomial. </li>

233 </ul><ul><li>So, it can be written as: \((\frac{x - 5}{2})^2 \)

234 </li>

235 <li>Hence, the expression becomes: \(x^2 - 5x = (\frac{x - 5}{2})^2 - \frac{25}{4} \)

236 </li>

237 </ul>Therefore, the answer is:

238 \( x^2 - 5x = (\frac{x - 5}{2})2 - \frac{25}{4} \)

239 Well explained 👍

240 <h2>FAQs on Completing the Square</h2>

241 <h3>1.What do you mean by completing the square?</h3>

242 In algebra, we use the method of completing the square to transform a quadratic expression, \(ax^2 + bx + c\), into a vertex form \(a (x + m)^2 + n\). It is useful for<a>solving quadratic equations</a>and for finding the minimum or maximum value of a quadratic expression.

243 <h3>2.Define a perfect square trinomial</h3>

244 It is a quadratic expression like \(a^2 + 2ab + b^2 \) or \(a^2 - 2ab + b^2\) that can be factored as \((a + b)^2\) or \((a - b)^2\). This expression results from squaring a<a>binomial</a>.

245 <h3>3.Can the coefficient of x2 be a number other than 1?</h3>

246 Yes, the coefficient of x2 can be any number other than 1 in a quadratic equation. If 1 is not the coefficient, divide the entire equation by that coefficient before the process. If 1 is the coefficient of x2, completing the square is easier.

247 <h3>4.What do we add to complete the square?</h3>

248 To complete the square in an expression \(ax^2 + bx + c\), we need to add and subtract \((\frac{b}{2a})^2\). This converts the quadratic expression into a perfect square trinomial. Thus, it results in \((x + \frac{b}{2a})^2 - (\frac{b}{2a})^2 + c\).

249 <h3>5.What is the purpose of adding and subtracting the same number?</h3>

250 In completing the square, we add and subtract the same number to keep the quadratic equation balanced. So that the value of the equation remains unchanged, without changing the original expression.

251 <h3>6.Why is completing the square important for building my child’s math skills?</h3>

252 It strengthens algebraic thinking, improves problem-solving abilities, and helps students tackle quadratic equations with confidence.

253 <h3>7.Can completing the square be used in subjects other than math?</h3>

254 Yes! It is used in physics for calculating projectile motion, in economics to find maximum profits, and in engineering for designing parabolic structures.

255 <h3>8.How can I help my child practice this topic at home?</h3>

256 Encourage step-by-step practice of different quadratic equations, use visual aids,<a>worksheets</a>, and online interactive videos to reinforce understanding.

257 <h2>Jaskaran Singh Saluja</h2>

258 <h3>About the Author</h3>

259 Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.

260 <h3>Fun Fact</h3>

261 : He loves to play the quiz with kids through algebra to make kids love it.