Table of Contents
Last modified on August 3rd, 2023
Completing the square is a way of rearranging quadratic equations from the general form ax2 + bx + c = 0 to the vertex form a(x – h)2 + k = 0. It is written as a(x + m)2 + n, such that the left side is a perfect square trinomial.
Comparing this equation with the general form, we get:
It is a form used to solve quadratic equations that is also useful in graphing and analyzing a quadratic function and finding its minimum or maximum values.
For example, the quadratic equation x2 – 4x + 4 = 0 can also be written as (x – 2)2. Although the two equations are equivalent, the form (x – 2)2 is sometimes better to work with in some situations.
Let us try to understand the concept using the concept of geometry.
In a quadratic equation ax2 + bx + c, we will arrange the expression in the form of a perfect square trinomial. Here, x comes twice, which makes it tough to solve. We will simplify the equation further using the concept of geometry as shown below:
We will start by dividing all terms by a, the coefficient of x2
Now, let us divide the equation into a square of side x and a rectangle of length ${\dfrac{b}{a}}$ and breadth x as shown:
Thus by adding ${\left( \dfrac{b}{2a}\right) ^{2}}$to the equation ${x^{2}+\dfrac{b}{a}x+\dfrac{c}{a}=0}$we can complete the square.
Step 1: Dividing all terms by a, the coefficient of x2
${x^{2}+\dfrac{b}{a}x+\dfrac{c}{a}=0}$
Step 2: Rewriting the equation such that the constant term ‘c’ is the only term on the right side of the equation
=>${x^{2}+\dfrac{b}{a}x=-\dfrac{c}{a}}$
Step 3:Adding the square of half the coefficient of the x -term, ${\left( \dfrac{b}{2a}\right) ^{2}}$ to both sides of the equation.
Step 4:Factoring the left side as the square of the binomial
=>${x^{2}+\dfrac{b}{a}x+\left( \dfrac{b}{2a}\right) ^{2}=-\dfrac{c}{a}+\left( \dfrac{b}{2a}\right) ^{2}}$
= ${\left( x+\dfrac{b}{2a}\right) ^{2}= -\dfrac{c}{a}+\dfrac{b}{( 2a}) ^{2}}$
This equation is in the vertex form a(x – h)2 + k. From here, we can find the vertex
Step 5: Taking the square root of both sides of the equation
=>${\sqrt{\left( x+\dfrac{b}{2a}\right) ^{2}}=\sqrt{-\dfrac{c}{a}+\begin{bmatrix} – \\ 20 \end{bmatrix}^{2}}}$
Step 6: Subtracting the number that remains on the left side of the equation to find x
=>${x+\dfrac{b}{2a}-\dfrac{b}{2a}=\sqrt{-\dfrac{c}{a}+\left( \dfrac{b}{2a}\right) ^{2}-}\dfrac{b}{2a}}$
=>${x=\sqrt{-\dfrac{c}{a}+\left( \dfrac{b}{2a}\right) ^{2}-}\dfrac{b}{2a}}$
By completing the square, let us solve the quadratic equation x2 + 4x + 1 = 0.
Step 1: Here, this step is not needed as a = 1
Step 2:Moving the constant term to the right side of the equation
=>x2 + 4x = -1
Step 3:: Adding the square of half the coefficient of the x -term to both sides of the equation.
${\left( \dfrac{4}{2}\right) ^{2}=\left( 2\right) ^{2}=4}$
=>x2 + 4x + 4 = -1 + 4
Step 4:Factoring the left side as the square of the binomial
=>(x + 2)2 = 3
This gives us the vertex of the parabola corresponding to the given equation as (-2, -3)
Step 5: Taking the square root on both sides of the equation
=>x + 2 = ±√3 = ±1.73
Step 6: Subtracting 2 from both sides of the equation
=> x = (-2 + √3, -2 – √3)
Find the value of c in the given quadratic equation x2 + 9x + c that completes the square.
Given the quadratic equation, x2 + 9x + c, here b = 9
As we know,
x2 + 9x + c
=> ${c=\left( \dfrac{b}{2}\right) ^{2}}$
=> ${c=\left( \dfrac{9}{2}\right) ^{2}}$
=> ${c=\dfrac{81}{4}}$
So, the value of c that makes the quadratic equation a perfect square trinomial is ${c=\dfrac{81}{4}}$
Solvex2 + 12x + 32 = 0by the method of completing the square
x2 + 12x + 32 = 0
=>x2 + 12x = -32
${\left( \dfrac{12}{2}\right) ^{2}=36}$
=>=> x2 + 12x + 36 = -32 + 36
=>(x + 6)2 = 4
=>${\sqrt{\left( x+6\right) ^{2}}=\sqrt{4}}$
=>x + 6 = ±2
=> x = (-4, -8)
Solve 2x2 – 4x – 5 = 0 by the method of completing the square
2x2 – 4x – 5 = 0
=>${x^{2}-2x-\dfrac{5}{2}=0}$
=>${x^{2}-2x=-\dfrac{5}{2}}$
${\left( \dfrac{1}{2}\times \left( -2\right) \right) ^{2}=1}$
=>${x^{2}-2x+1=\dfrac{5}{2}+1}$
=>${\left( x-1\right) ^{2}=\dfrac{7}{2}}$
=>${\sqrt{\left( x-1\right) ^{2}}=\pm \sqrt{\dfrac{7}{2}}}$
=>${\sqrt{\left( x-1\right) ^{2}}=\pm \sqrt{\dfrac{7}{2}}\times \dfrac{\sqrt{2}}{\sqrt{2}}=\pm \dfrac{\sqrt{14}}{2}}$
=>${x-1=\pm \dfrac{\sqrt{14}}{2}}$
=>${x=1\pm \dfrac{\sqrt{14}}{2}}$
=>${x=\dfrac{2}{2}\pm \dfrac{\sqrt{14}}{2}}$
=>${x=\left( \dfrac{2+\sqrt{14}}{2},\dfrac{2-\sqrt{14}}{2}\right)}$
Solving a quadratic equation of the form a(x + m)2 + n, where a = 1
Expanding (x + m)2 + n, we get x2 + 2mx + m2 + n
Now, if we compare a quadratic equation of the form ax2 + bx + c with the above equation, we will obtain the value of m and n
Let us use the equation x2 + 12x + 32 = 0
Comparing this equation with the equation x2 + 2mx + m2 + n, we get
2mx = 12x
=>m = 6
m2 + n = 32
=> (6)2 + n = 32
=> 36 + n = 32
=> n = -4 Thus, we get the same result (x + 6)2 – 4 as obtained while solving the quadratic equation in solved example 1. The rest of the steps remains the same while solving the quadratic equation, and we will get the result x = (-4, -8)
Last modified on August 3rd, 2023