Solving Quadratic Equations

A quadratic equation, typically in the form ax² + bx + c = 0, can be solved using different methods including factoring, completing the square, quadratic formula, and the graph method. While Solving Quadratic Equations we try to find a solution that represent the points where this the condition Q(x) = 0. The solutions are also called roots or zeros of the quadratic equation.

Quadratic Equation

Quadratic Equation can be defined as a polynomial equation of a second degree, which implies that it includes at least one term that is squared.

ax² + bx + c = 0 

a, b and c are real numbers while a ≠ 0. Its shape is a parabola that opens upwards or downwards depending upon the value of "a". 

Its solution is the value of x for which the equation is satisfied or the point on the cartesian plane where the graph of quadratic equation intersects the x- axis.

Methods of Solving Quadratic Equation

Completing Square Method  

In the completing Square method of Solving Quadratic Equation need to bring the equation in the form of whole squares, for example: (x - a)² - b² = 0. 

Steps for finding out roots by completing the square method:

Step 1: Bring the equation in the form ax² + bx = -c.

Step 2: We need to make sure that a = 1 (if a≠1, multiply through the equation by \frac{1}{a} before going to next step.)

Step 3: Use the value of b from this new equation and (\frac{b}{2})^2 to both sides of the equation to form a perfect square on the left side of the equation.

Step 4: Find the square root of the both sides of the equation.

Step 5: Solve the result to get the roots.

Example: . Find the roots of the equation by completing the square's method of Solving quadratic Equation.  9x² + 24x + 3 = 0 
Solution:

9x² + 24x + 3 = 0 

This equation can be re-written as, 

⇒ 9x² + 24x + 16 - 13 = 0 
⇒ (3x)² + 24x + 4² -13 =0 
⇒ (3x + 4)² -13 = 0 
⇒ (3x + 4)² -(√13)² = 0 
⇒ (3x + 4)² = (√13)²

Taking square root of the both sides of the equation. 
3x + 4 = √13 or 3x + 4 = -√13

We get our roots by solving these two equations, 

3x + 4 = √13 
x =  \frac{√13 - 4}{3} 

Similarly,

3x + 4 = - √13 
x =  \frac{-√13 - 4}{3} 

Factoring Method

In the factorization method of Solving Quadratic Equation we try to factor out the equation such that we get the equation in form of the product of two terms. Then on equating these two terms to zero, we get the roots. 

Step 1: Write the quadratic equation in standard form

Step 2: Find two numbers such that the product of the numbers is 'ac' and the sum is 'b'.

Step 3: Then write x coefficient as the sum of these two numbers and split them such that you get two terms for x.

Step 4: Factor the first two as a group and the last two terms as another group.

Step 5: Take common factors from these and on equating the two expressions with zero after taking common factors and rearranging the equation we get the roots.

Example: Factorize the following equation and find its roots: 2x² - x - 1 = 0.
Solution:

2x² - x - 1 = 0 

⇒ 2x² -2x + x - 1 = 0 
⇒ 2x(x - 1) + 1(x - 1) = 0 
⇒ (2x + 1) (x - 1) = 0 

For this equation two be zero, either one of these or both of these terms should be zero. 
So, we can find out roots by equating these terms with zero. 

2x + 1 = 0
x = -\frac{1}{2}   
x - 1 = 0 
⇒ x = 1 

So, we get two roots in the equation. 
x = 1 and -\frac{1}{2}

Quadratic Formula

The quadratic formula provides a straightforward way to find the solutions and it work on all quadratic equations. In quadratic formula, a is the coefficient of the quadratic term x², b is the coefficient of the linear term x, and c is the constant term. The symbol ± indicates that there are typically two possible solutions: one for the positive sign and one for the negative sign.

Quadratic formula to solve the general quadratic equation ax² + bx + c = 0 is given in the following image:

Nature of Roots

We can easily find the nature of roots of the quadratic equations without actually finding the roots. Generally, we represent the roots with α and β symbols.

Let's understand the nature of roots and the corresponding value of discriminant related to it.

Example : Find out the roots of the equation using Quadratic Formula,  4x² + 10x + 3 = 0
Solution:

4x² + 10x + 3 = 0

Using Quadratic Formula to solve this,
a = 4, b = 10 and c = 3

Before plugging in the values, we need to check for the discriminator 
b² - 4ac 

⇒ 10² - 4(4)(3) 
⇒ 100 - 48 
⇒ 52 

This is greater than zero, So now we can apply the quadratic formula. 
Plugging the values into quadratic equation, 

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ = \frac{-10 \pm \sqrt{10^2 - 4(4)(3)}}{2(4)} \\ = \frac{-10 \pm \sqrt{100 - 48}}{8} \\ = \frac{-10 \pm \sqrt{52}}{8}

Graph Method

The graphical method for solving quadratic equations involves plotting the quadratic function and identifying the points where the graph intersects the x-axis. These intersections represent the roots of the quadratic equation.

Let us suppose the general form of the quadratic equation is ax² + bx + c = 0, where a ≠ 0. The quadratic equation is a polynomial equation of degree 2, so it comes under the conic section.

Further simplifying the standard form of quadratic equation,

 y = ax² + bx + c 

⇒ y = a[(x + b/2a)² – (D/4a²)]

⇒ y - D/4a = a[(x + b/2a)²]

Now this equation clearly resembles a parabola and we can easily draw its curve. The points where this curve cut the x-axis are the roots of the quadratic equation (or zeroes of the quadratic polynomial).

Solved Examples on Solving Quadratic Equations

Question 1: Solve the quadratic equation x2+6x−7=0x^2 + 6x - 7 = 0x2+6x−7=0 by completing the square.
Solution:

Given equation: x² + 6x - 7 = 0

Move the constant term to the right side:
x²+ 6x = 7.

Complete the square by adding the square of half the coefficient of x on both sides i,e (6/2)² = 9.
x² + 6x + 9 = 7 + 9

which simplifies to:
(x + 3)² = 16

Solve for x by taking the square root of both sides:
x + 3 = ± 4

Solve for x:
x + 3 = 4 gives x = 1,
x + 3 = −4 gives x = −7.

So, the solutions are x = 1 and x = −7.

Question 2: Solve the quadratic equation x² + 5x + 6 = 0 by factoring the expression.
Solution:

Given equation: x² + 5x + 6 = 0.

Factor the quadratic expression:
Find two numbers that multiply to 6 (the constant term) and add to 5 (the coefficient of x). The numbers are 2 and 3, since 2 × 3 = 6 and 2 + 3 = 5.

Factor the quadratic expression as:
(x + 2)(x + 3) = 0.

Solve for x by setting each factor equal to zero:
x + 2 = 0 gives x = −2,
x + 3 = 0 gives x = −3.

So, the solutions are x = −2 and x = −3.

Question 3: Solve the quadratic equation 2x²−3x−5=0 using the quadratic formula.
Solution:

Given equation: 2x² − 3x − 5 = 0.

The quadratic formula is given by : x=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a} where a = 2, b = -3, and c = −5.

Substitute the values of a, b, and c into the formula : x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-5)}}{2(2)}

Simplifying : x = \frac{3 \pm \sqrt{9 + 40}}{4}

x = \frac{3 \pm \sqrt{49}}{4}
x = 4\frac{3 \pm 7}{4}

Solve for x:

x = = \frac{3 + 7}{4} = \frac{10}{4} = 2.5
x = \frac{3 - 7}{4} = \frac{-4}{4} = -1

So, the solutions are x = 2.5 and x = −1.

Also Read, Roots of quadratic Equations