So far, you’ve learned how to factorize special cases of quadratic equations using the difference of square and perfect square trinomial method. Show These methods are relatively simple and efficient; however, they are not always applicable to all quadratic equations. In this article, we will learn how to solve all types of quadratic equations using a simple method known as completing the square. But before that, let’s have an overview of the quadratic equations. A quadratic equation is a polynomial of second degree, usually in the form of f(x) = ax2 + bx + c where a, b, c, ∈ R, and a ≠ 0. The term ‘a’ is referred to as the leading coefficient, while ‘c’ is the absolute term of f (x). Every quadratic equation has two values of the unknown variable, usually known as the roots of the equation (α, β). We can obtain the root of a quadratic equation by factoring the equation. What is Completing the Square?Completing the square is a method of solving quadratic equations that we cannot factorize. Completing the square means manipulating the form of the equation so that the left side of the equation is a perfect square trinomial. How to Complete the Square?To solve a quadratic equation; ax2 + bx + c = 0 by completing the square. The following are the procedures:
⟹ (b/2a)2.
x + q= ± √r
Complete the square formulaIn mathematics, completing the square is used to compute quadratic polynomials. Completing the Square Formula is given as: ax2 + bx + c ⇒ (x + p)2 + constant. The quadratic formula is derived using a method of completing the square. Let’s see. Given a quadratic equation ax2 + bx + c = 0; Isolate the term c to right side of the equation ax2 + bx = -c Divide each term by a. x2 + bx/a = -c/a Write as a perfect square (x + b/2a) 2= (-4ac+b2)/4a2 (x + b/2a) = ±√ (-4ac+b2)/2a x = – b/2a ±√ (b2– 4ac)/2a x = [- b ±√ (b2– 4ac)]/2a………. (This is the quadratic formula) Now let’s solve a couple of quadratic equations using the completing square method. Example 1 Solve the following quadrating equation by completing square method: x2 + 6x – 2 = 0 Solution Transform the equation x2 + 6x – 2 = 0 to (x + 3)2 – 11 = 0 Since (x + 3)2 =11 x + 3 = +√11 or x + 3 = -√11 x = -3+√11 OR x = -3 -√11 But √11 =3.317 Therefore, x = -3 +3.317 or x = -3 -3.317, x = 0.317 or x = -6.317 Example 2 Solve by completing square x2 + 4x – 5 = 0 Solution The standard form of completing square is; In this case, b = 4, c = -5. Substitute the values; Example 3 Solve x2 + 10x − 4 = 0 Solution Rewrite the quadratic equation by isolating c on the right side. x2 + 10x = 4 Add both sides of the equation by (10/2)2 = 52 = 25. = x2 + 10x + 25 = 4 + 25 = x2 + 10x + 25 = 29 Write the left side as a square (x + 5) 2 = 29 x = -5 ±√29 x = 0.3852, – 10.3852 Example 4 Solve 3x2 – 5x + 2 = 0 Solution Divide each term of the equation by 3 to make the leading coefficient equals to 1. Example 5 Solve x2 – 6x – 3 = 0 Solution x2 – 6x = 3 (x – 3)2 = 12 x – 3= ± √12 x = 3 ± 2√3 Example 6 Solve: 7x2 − 8x + 3=0 Solution 7x2 − 8x = −3 x2 −8x/7 = −3/7 x2 – 8x/7 +(−4/7)2 = −3/7+16/49 (x − 4/7)2 = −5/49 x = 4/7 ± (√7) i/5 (x – 3)2 = 12 x − 3 = ±√12 x = 3 ± 2√3 Example 7 Solve 2x2 – 5x + 2 = 0 Solution Divide each term by 2 x2 – 5x/2 + 1 = 0 ⇒ x2 – 5x/2= -1 Add (1/2 × −5/2) = 25/16 to both sides of the equation. = x2 – 5x/2 + 25/16 = -1 + 25/16 = (x – 5/4)2 = 9/16 = (x – 5/4)2 = (3/4)2 ⇒ x – 5/4= ± 3/4 ⇒ x = 5/4 ± 3/4 x = 1/2, 2 Example 8 Solve x2– 10x -11= 0 Solution Write the trinomial as a perfect
square ⇒ (x – 5)2 – 36 =0 ⇒ (x – 5)2 = 36 Find the square roots on both sides of the equation x – 5 = ± √36 x -5 = ±6 x = −1 or x =11 Example 9 Solve the following equation by completing the square x2 + 10x – 2 = 0 Solution x2 + 10x – 2 = 0 ⇒ x2 + 10x = 2 ⇒ x2 + 10x + 25 = 2 + 25 ⇒ (x + 5)2 = 27 Find the square roots on both sides of the equation ⇒ x + 5 = ± √27 ⇒ x + 5 = ± 3√3 x = -5 ± 3√3 Example 10 Solve x2 + 4x + 3 = 0 Solution x2 + 4x + 3 = 0 ⇒ x2 + 4x = -3 x2 + 4x + 4 = – 3 + 4 Write the trinomial as a perfect square (x + 2)2 = 1 Determine the square roots on both sides. (x + 2) = ± √1 x= -2+1= -1 OR x = -2-1= -3 Example 11 Solve the equation below using the method of completing the square. 2x2 – 5x + 1 = 0 Solution x2−5x/2 + 1/2=0 x2 −5x/2 = −1/2 (1/2) (−5/2) =−5/4 (−5/4)2 = 25/16 x2 − 5x/2 + 25/16 = −1/2 + 25/16 (x – 5/4) 2 = 17/16 Find the square of both sides. (x – 5/4) = ± √ (17/16) x = [5 ± √ (17)]/4 How do you complete the square step by step?Completing The Square Steps
Isolate the number or variable c to the right side of the equation. Divide all terms by a (the coefficient of x2, unless x2 has no coefficient). Divide coefficient b by two and then square it. Add this value to both sides of the equation.
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