June 13, 2024

What are 5 methods of Solving a Quadratic Equation ?

What are 5 methods of Solving a Quadratic Equation ?

A quadratic equation is a polynomial equation in a single variable of the form ax^2+bx+c=0

 where x is a  variable and a, b\, \, and \, \, c are constants. The term ax^2 is the quadratic term, bx is the linear term and c is the constant term.
A quadratic equation has two solutions. Either two distinct real solutions, one double real solution or two imaginary solutions. There are several methods you can use to solve a quadratic equation.

5 methods of Solving a Quadratic Equation

(i) Factoring(split the middle term)
(ii)Principle of square roots
(iii)Completing the Square
(iv) Quadratic Formula
(v) Graphing

Factoring(split the middle term): 

Factoring a quadratic equation by splitting the middle term involves breaking the middle term of the quadratic expression into two terms, which then allows you to factor by grouping. The steps are as follows:
1. Start with a quadratic equation in the form ax^2+bx+c=0, where a, b \, \, and\, \, c are constants. Example:x^2+5x+6=0
2. Identifya, b \, and \, c: In the example x^2+5x+6=0, a=1, b=5, and \, c=6.
3.  Split the middle term ( bx) into two terms whose coefficients multiply to give ac(the product of
a \, \, and \, \, c). For x^2+5x+6, you want two numbers whose product is 1 \times 6 =6 and whose sum is 5. The numbers 2 and fit these criteria, so you can split 5x into 2x+3x.
x^2+5x+6=x^2+2x+3x+6=0
4. Group the terms in pairs:(x^2+2x)+(3x+6)=0
5. Factor out the common factor from each group:
x(x+2)+3(x+2)=0
6. Factor out the common binomial factor:
(x+2)(x+3)=0
7. Set each factor equal to zero and solve forxx+2=0 \, \, or\, \, x+3=0
So, the solutions to the quadratic equationx^2+5x+6=0 are x=-2 \, or\, x=-3

Principle of square roots :

The square roots method is one of the methods used to solve quadratic equations. The general form of a quadratic equation is:
ax^2+bx+c=0,
The square roots method is applicable when the quadratic equation can be expressed in the form:
(x-p)^{2}=q where p and  q are constants.
To solve for x, you take the square root of both sides of the equation and then isolate x:
x= p\pm \sqrt{q}
This gives you two possible solutions for the quadratic equation. The \pmsymbol indicates that there are generally two solutions, one with the positive square root and the other with the negative square root.

It’s important to note that not all quadratic equations can be easily factored into the form (x-p)^2=q, and in such cases, other methods like factoring, completing the square, or using the quadratic formula may be more appropriate.

Example : x^2-16=0
Solution:
Moving the constant to the right side , we get x^2=16
Take the square root of both sides \sqrt{x^2} =\pm \sqrt{16}
\Rightarrow x=\pm 4
So the solution of given equation  is x= 4 \, or \, x=-4 .

Example :  2(x+3)^{2}-14=0
Solution : 2(x+3)^{2}-14=0 \Rightarrow 2(x+3)^2=14
\Rightarrow (x+3)^2=7
\Rightarrow \sqrt{(x+3)^2}=\sqrt{7}
\Rightarrow x+3= \pm \sqrt{7}
\Rightarrow x=-3+\sqrt{7} \, or\, x=-3-\sqrt{7}

Completing the Square method of solving quadratic equation:

Here’s a step-by-step guide on how to use the completing the square method:
1. Start with the quadratic equation  ax^2+bx+c=0.
2. If a\neq 1, divide the entire equation by a to make the coefficient of x^2 equal to 1.
x^2+\frac{b}{a}x+\frac{c}{a}=0
3. Rearrange the equation to isolate the x^2 and undefinedx terms on one side and move the constant term to the other side.
x^2+\frac{b}{a}x=-\frac{c}{a}
4. Add and subtract  (\frac{b}{2a})^{2}

x^2+\frac{b}{a}x+\left ( \frac{b}{2a} \right )^2=-\frac{c}{a}+\left ( \frac{b}{2a} \right )^25. Factor the left side into a perfect square trinomial.
\left ( x+\frac{b}{2a} \right )^{2}=-\frac{c}{a}+\left ( \frac{b}{2a} \right )^2
6. Take the square root of both sides and solve for x.
x+\frac{b}{2a}=\pm \sqrt{-\frac{c}{a}+\left ( \frac{b}{2a} \right )^2}
\Rightarrow x=-\frac{b}{2a}\pm \sqrt{-\frac{c}{a}+\left ( \frac{b}{2a} \right )^2}

This is the solution to the quadratic equation using the completing the square method. Note that sometimes the square root term simplifies to a real number, while other times it results in complex numbers.

Example : solve the quadratic equation 2x^2-4x-3=0using the completing the square method.
Solution :
Given that  2x^2-4x-3=0.
Since a=2, divide the entire equation by 2. We get x^2-2x-\frac{3}{2}=0
Rearrange the equation to isolate the x^2 and x terms on one side and move the constant term to the other side.
x^2-2x=\frac{3}{2}. Now add and subtract (\frac{b}{2a})^{2}= \left ( \frac{-2}{2\times 2} \right )^2=1
x^2-2x+1=\frac{3}{2}+1
\Rightarrow (x-1)^2=\frac{5}{2}
\Rightarrow x-1=\pm \sqrt{\frac{5}{2}}
\Rightarrow x=1\pm \sqrt{\frac{5}{2}}
So, the solutions to the quadratic equation 2x^2-4x-3=0are

Quadratic Formula for solving a quadratic equation

Consider a quadratic equation in the standard form ax^2+bx+c=0, where a,b and c are constants and a\neq 0.
The quadratic formula gives the solutions for  x in terms of a,b and c. The formula is

x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

The expression under the square root, b^2-4ac is called the discriminant (\Delta).

  • If \Delta >0, the quadratic equation has two distinct real roots.
  • If \Delta =0, the quadratic equation has one real root (the parabola touches the x-axis at one point).
  • If \Delta <0 the quadratic equation has two complex roots (conjugate pairs).

Example : Find the roots of the equation 2x^2-5x+2=0 .
Here a=2 , b=-5 and c=2.
Now, let’s apply the quadratic formula:

x=\frac{-(-5)\pm\sqrt{(-5)^2-4\times 2\times 2}}{2\times 2}

=\frac{-(-5)\pm\sqrt{25-16}}{4}

=\frac{5\pm\sqrt{9}}{4}

=\frac{5\pm\3}{4}

So the solutions are x_1=\frac{5+3}{4} =\frac{8}{4}=2 and x_2=\frac{5-3}{4}=\frac{2}{4}=\frac{1}{2} .

Therefore the solutions are x=2  and x=\frac{1}{2} .

Graphing method for solving a quadratic equation

 For any quadratic polynomial , ax^{2}+bx+c , a\neq 0 the graph of the corresponding equation  ax^{2}+bx+c   has one of the two shapes either open upwards like U or open   downwards like ,   ∩   depending on whether a > 0 or a < 0. (These curves are called parabolas.) 

Graph-of-a-quadratic-function

The solutions of the equation are the 𝑥 values for which the function is zero, which we refer to as the  roots of the function. On a graph, these values are the 𝑥-coordinates of the points where the 𝑦-value is zero, which corresponds to the points at which the graph crosses the 𝑥-axis.
 A quadratic equation will have up to two real solutions. If an equation has two solutions, the corresponding function will have a graph that crosses the 𝑥-axis twice.  An equation with a repeated solution will lead to a graph that has a vertex on the 𝑥-axis. Finally, an equation having no solution will mean that the graph is entirely above or below the 𝑥-axis.
Roots of a quadratic equation
In the graphs above, the first function has two roots, the middle function has one root where the graph touches the 𝑥-axis, and the last function has no roots.

In order to find the solutions of a quadratic equation using a graph:
(i) Rearrange the equation so that one side  (if necessary).
(ii) Draw the graph of the quadratic function.
(iii) Read off the -coordinate(s) of the point(s) where the curve crosses the -axis.

Example :  Find the solutions of the equation x^2+6x+9=0 graphically.
Solution: Draw the graph of x^2+6x+9=0 .

Now read off the -coordinate(s) of the point(s) where the curve crosses the -axis.
Here we have one (repeated) root at x=-3 . The solution to the equation x^2+6x+9=0 is  x=-3 .

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