Finding roots using the Quadratic Formula--Conceptmathhelp.com

What is the Quadratic Formula?

The Quadratic formula is one of the methods to find the roots of a quadratic equation.

It is a faster approach to find the roots of a quadratic equation when compared to, Factorization method and Completing the square method.

How to use quadratic formula?

To find the roots of a quadratic equation using Quadratic formula, all we need is to compare the given quadratic with the standard form, get the coefficients a,b,c and lastly need to plug into the quadratic formula and simplify.

Unlike the Factorization approach, there will be no tricky steps, which involve the splitting of the middle term into two parts P and Q ,so as to get the sum of the two parts P and Q as the middle term and the product of two terms P and Q as the product of first and last terms of the quadratic equation.

Unlike Completing the square approach, no efforts are needed to make the leading coefficient 1 by dividing the equation throughout with the leading coefficient “a” and finding the constant number thereby which helps in writing a perfect square quadratic.

How does the formula look like?

The quadratic formula as shown below.

Quadratic formula and the Discriminant — Quadratic formula and The Discriminant

What is the Discriminant of a Quadratic equation?

The Expression that is present under the radicle in the Quadratic formula $b^{2}-4ac$ is called as the discriminant of the quadratic equation.

Discriminant and nature of roots of a Quadratic Equation

Roots of a quadratic equation can be classified into 2 types,

Real Roots
Complex roots.

Real Roots:-

If the graph of the Quadratic Equation intersects or touches the “x” axis, then we can conclude that the roots of the given quadratic will be Real.

Real Roots can be in turn classified into

i) Real and distinct roots

ii) Real and Equal roots.

Complex roots:-

If the Graph of the Quadratic Equation does not intersect the “x”- Axis, then the quadratic equation will not have Realroots.It will have the complex roots.

How can the Discriminant be used to find the nature of roots?

If the Discriminant $b^{2}-4ac\geq 0$ , Then the roots of the quadratic equation will be Real.
If $b^{2}-4ac=0$ , the roots would be real and equal.
If $b^{2}-4ac>0$ the roots would be real and distinct.
If the Discriminant $b^{2}-4ac <0$ , then the roots of the Quadratic would be Complex or Imaginary.

What is the relation between the Graph of a Quadratic and its Discriminant?

The number of x-intercepts of a quadratic graph depends on the number of real roots. a quadratic

The number of real roots of a quadratic is decided by the Discriminant.

So discriminant of a quadratic conveys how many “x”- intercepts a quadratic equation can have.

If the Discriminant $b^{2}-4ac=0$ , then the Graph of the Quadratic intersects “x”-axis at only one point. Such a quadratic will have a single solution with repeated root.It is also called as a double root.
If the Discriminant $b^{2}-4ac >0$ , then the graph of the Quadratic intersects “x”-axis at two different points. Such a Quadratic equation is said to have two distinct roots.
If the Discriminant $b^{2}-4ac <0$ , then the graph of the quadratic will NOT intersect the “x”-axis and the quadratic is said to have no solutions.

What are the common mistakes committed while using Quadratic formula?

Though the Quadratic formula is easy to use, we need to pay Close attention while implementing.Following are the common mistakes we observed while implementation of the formula

The division 2a is for the complete numerator $-b\pm \sqrt {b^{2}-4ac}$ .Avoid writing $-b\pm \dfrac {\sqrt {b^{2}-4ac}}{2a}$ or $-\dfrac {b}{2a}\pm \sqrt {b^{2}-4ac}$
The $\pm$ before $\sqrt {b^{2}-4ac}$ must not be ignored.
To avoid creeping of errors when the coefficients are negative value.Common tendency to write $\sqrt {-b^{2}-4ac}$ .Instead ,make it a practice fo writing $\sqrt {\left( -b\right) ^{2}-4ac}$
When the coefficients “a “or “c” are negative,the discriminant would be $b^{2}+4ac$ .It is always a good practice to write the formula in parenthesis as shown $\dfrac {-b\pm \sqrt {b^{2}-4\left( -a\right) \left( c\right) }}{2a}$ to aviod writing $b^{2}-4ac$ .

Summary and points to be noted:

Quadratic formula $x=\dfrac {-b\pm \sqrt {b^{2}-4ac}}{2a}$ can be applied to any Quadratic equation to get quick insight about the X-intercepts, Nature of roots whether they are real -distinct Or Real -equal or Complex roots.
Pay close attention when writing the denominator part 2a, It is a common denominator.Also be cautious when the quadratic equation has negative coefficients a,b,c.
Discriminant $b^{2}-4ac$ does the job of deciding the nature of roots. $b^{2}-4ac=0 then --Real-equal -- two "x" -intercepts \\b^{2}-4ac=>0 --Real-Distinct --single "x"- intercept\\b^{2}-4ac<0 --Complex-Rotos-- No "x"- intercepts.$

Worked out examples

Let us start with a simple example.

$x^{2}-5x+6$

Strategy Step-by-Step:-

Step1: Compare the given quadratic with the standard form of the quadratic equation $y= ax^{2}+bx +c$

On one to one comparison with the standard form the quadratic equation $y= ax^{2}+bx +c$ ,the leading a=1 ,b= -5,c=6.

Step2: Plug in these values into the quadratic formula $x=\dfrac {-b\pm \sqrt {b^{2}-4ac}}{2a}$

$x=\dfrac {-(-5)\pm \sqrt {(-5)^{2}-4(1)(6)}}{2(1)}$

$\begin{aligned}\Rightarrow \dfrac {+5\pm \sqrt {25-4\times 6}}{2}\\ \Rightarrow \dfrac {5\pm \sqrt {1}}{2}\end{aligned}\\ \begin{aligned}\Rightarrow \dfrac {5+1}{2},\dfrac {5-1}{2}\\ \Rightarrow \dfrac {6}{2},\dfrac {4}{2}\\ \Rightarrow 3,2\end{aligned}$

Step3:Simiplify the expression and find the value of the roots keeping in mind the frequent mistakes that can be done.

Simplifying further

$\dfrac {+5\pm \sqrt {25-4\times 6}}{2}\\ \Rightarrow \dfrac {5\pm \sqrt {1}}{2}\end{aligned}\\\begin{aligned}\Rightarrow \dfrac {5+1}{2},\dfrac {5-1}{2}\\ \Rightarrow \dfrac {6}{2},\dfrac {4}{2}\\ \Rightarrow x= 3,2\end{aligned}$

Therefore the roots of the quadratic would be 3 and 2.

Derivation of Quadratic formula.

Quadratic formula can be derived from the “completing the square method” for finding the roots of a Quadratic Equation.

In fact, the Quadratic formula is the end product of “completing the square method.”

Let us derive the quadratic formula, using the steps of “completing the square” method.

The standard form of a quadratic equation is given $y= ax^{2}+bx +c$ .

Step1: As the first step, replace y with 0.Identify the leading coefficient, which is the coefficient of $x^{2}$

The standard form of a quadratic equation, $y= ax^{2}+bx +c$ .Replacing “y” with 0 we will have $0=ax^{2}+bx +c$ .The leading coefficient is “a” for the standard form

Step2: Divide the equation throughout by the leading coefficient “a” to make the coefficient of $x^{2}$ as 1.

Therefore, on division with “a” both sides, we get

$\dfrac {ax^{2}+bx+c}{a}=\dfrac {0}{a}\\ \begin{aligned}\dfrac {ax^{2}}{a}+\dfrac {bx}{a}+\dfrac {c}{a}=0\\ \Rightarrow x^{2}+\dfrac {b}{a}x+\dfrac {c}{a}=0\end{aligned}$ .

Now we have the coefficient of $x^{2}$ as 1.

Step3: Find the value of the coefficient of “x” which is “b” of the new equation. Subsequently, divide the coefficient by 2 and square it.

The coefficient of “x” is $\dfrac {b}{a}$ .On division by 2 and squaring we get $\left( \dfrac {\dfrac {b}{a}}{2}\right) ^{2}=\left( \dfrac {b}{2a}\right) ^{2}$ .

Step4: Add and subtract the term $\left( \dfrac {b}{2}\right) ^{2}$ term to the quadratic equation.

Therefore we get $x^{2}+\dfrac {b}{a}x+\dfrac {c}{a}+\left( \dfrac {b}{2a}\right) ^{2}-\left( \dfrac {b}{2a}\right) ^{2}=0$

Step5: Group the First term, Middle term and + $\left( \dfrac {b}{2a}\right) ^{2}$ term to complete the square

Grouping the first, middle and + $\left( \dfrac {b}{2a}\right) ^{2}$ terms ,we then have $\left[ x^{2}+\dfrac {b}{a}x+\left( \dfrac {b}{2a}\right) ^{2}\right] -\left( \dfrac {b}{2a}\right) ^{2}+\dfrac {c}{a}=0$ .

Step6: Write the Completed square as $\left( x+\dfrac {b}{2}\right) ^{2}$ .Take the rest of the terms to the right side of “=” signand solve for the variable “x” taking square root both sides.

Completing the square, we will then have

$\left[ x+\dfrac {b}{2a}\right] ^{2}-\left( \dfrac {b}{2a}\right) ^{2}+\dfrac {c}{a}=0 \Rightarrow \left[ x+\dfrac {b}{2a}\right] ^{2}=\left( \dfrac {b}{2a}\right) ^{2}-\dfrac {c}{a} \begin{aligned}\Rightarrow \left[ x+\dfrac {b}{2a}\right] ^{2}=\dfrac {b^{2}}{4a^{2}}-\dfrac {c}{a} \Rightarrow \left[ x+\dfrac {b}{2a}\right] ^{2}=\dfrac {b^{2}-4ac}{4a^{2}}\end{aligned}$

Taking Square root both sides

$\begin{aligned}\Rightarrow \sqrt {\left( x+\dfrac {b}{2a}\right) ^{2}}=\pm \sqrt {\dfrac {b^{2}-4ac}{4a^{2}}}\\ \Rightarrow x+\dfrac {b}{2a}=\pm \dfrac {\sqrt {b^{2}-4ac}}{2a}\end{aligned}$

Isolating “x” by subtracting $\dfrac {b}{2a}$ both sides.

$\begin{aligned}\Rightarrow x=\dfrac {-b}{2a}\pm \dfrac {\sqrt {b^{2}-4ac}}{2a}\\ \Rightarrow x=\dfrac {-b\pm \sqrt {b^{2}-4ac}}{2a}\end{aligned}$ .

This is the Quadratic formula derived using completing the square method.