Example Of Solving A Quadratic Equation Using The Quadratic Formula The Theory of Quadratic Equations

You are searching about Example Of Solving A Quadratic Equation Using The Quadratic Formula, today we will share with you article about Example Of Solving A Quadratic Equation Using The Quadratic Formula was compiled and edited by our team from many sources on the internet. Hope this article on the topic Example Of Solving A Quadratic Equation Using The Quadratic Formula is useful to you.

The Theory of Quadratic Equations

A quadratic equation is a second order polynomial equation. A quadratic equation has two roots. Roots can also be equal and equal. Let’s write the quadratic equation in two forms

AX * X + BX + C = 0 An example of a quadratic equation would be 5X*X + 3 *X + 2 = 0.

Let’s rewrite the quadratic equation as (X-R1) * (X-R2) = 0. The above step is called factoring.

Let us also rewrite the original generalized equation of the quadratic as X*X + B/A * X + C/A = 0.

The qualitative equation can be rewritten as X * X -X (R1 + R2) + R1R2 = 0.

Comparing like terms we can see that -(R1 + R2) = B/A

R1R2 = C/A

(R1 + R2) = -B/A

Let’s investigate B*B – 4 * A * C

B = -A (r1 + r2)

C = AR1R2; 4*A*C = 4*A*A*R1*R2

B*B = A*A(R1 + R2) * (R1 + R2)

Differentiation = A*A(R1 + R2) * (R1 + R2) – 4*A*A*R1*R2

= A*A ( (R1+R2)((R1+R2) – 4R1R2)

= A*A (R1 – R2) * (R1 – R2).

Note that this is a perfect square of A(R1-R2). So if the difference is negative it means that the quadratic equation has no real roots because the squares of real numbers are also perfect squares.

Let A(R1-R2) be added to -B which is A(R1 + R2), and the sum is 2AR1. Dividing this by 2A will give R1.

Similarly subtract A(R1-R2) from -B ie, A(R1 + R2) – A (R1-R2)

which is equal to A(2R2) or 2AR2. Dividing this by 2A will give R2.

So R1 is (-B + square root (discriminant) ) / 2A and R2 is (-B – square root (discriminant) / 2A

Let’s take a look at some common factoring problems you’ll encounter

Say x * x + 5*x + 6 = 0.

Evaluate the first step discriminant which is equal to SQUAREROOT(25 – 24) = 1, which means there are real roots.

The roots of the equation are (- 5 + 1)/ 2 equal to -2 and ( -5 -1)/2 equal to -3.

The equation can be multiplied as (X+2)(X+3) = 0.

Let’s take another example

3 * x * x + 9 * x + 6 = 0, rewriting it as x * x + 3*x + 2 = 0.

discriminant = sqrt(9-8) = 1

R1 = -1 and R2 is -2. So the same equation is a multiplicative form

(x + 1)(x + 2) = 0.

A quadratic equation can also be plotted on a graph. When plotted this will produce the equation of a parabola.

Video about Example Of Solving A Quadratic Equation Using The Quadratic Formula

You can see more content about Example Of Solving A Quadratic Equation Using The Quadratic Formula on our youtube channel: Click Here

Question about Example Of Solving A Quadratic Equation Using The Quadratic Formula

If you have any questions about Example Of Solving A Quadratic Equation Using The Quadratic Formula, please let us know, all your questions or suggestions will help us improve in the following articles!

The article Example Of Solving A Quadratic Equation Using The Quadratic Formula was compiled by me and my team from many sources. If you find the article Example Of Solving A Quadratic Equation Using The Quadratic Formula helpful to you, please support the team Like or Share!

Rate Articles Example Of Solving A Quadratic Equation Using The Quadratic Formula

Rate: 4-5 stars
Ratings: 1158
Views: 1036749 5

Search keywords Example Of Solving A Quadratic Equation Using The Quadratic Formula

Example Of Solving A Quadratic Equation Using The Quadratic Formula
way Example Of Solving A Quadratic Equation Using The Quadratic Formula
tutorial Example Of Solving A Quadratic Equation Using The Quadratic Formula
Example Of Solving A Quadratic Equation Using The Quadratic Formula free
#Theory #Quadratic #Equations

Source: https://ezinearticles.com/?The-Theory-of-Quadratic-Equations&id=7557712