Quadratic Equations

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A quadratic equation is an equation written in the form of:

 

a{x}^{2}+bx+c=0

 

where a, b and c are real numbers with a not equal to 0. A quadratic equation is also called a second-degree polynomial equation in x.

Solving quadratic equations may produce three possible outcomes: no solution, one solution or two solutions. These techniques are also a part of calculus, in which the solutions are the roots of a parabola crossing the x-axis.

There are a number of simple quadratic equations that are straightforward to solve, such as:

{ x }^{ 2 }=8\\ \\ x=\pm \sqrt { 8 } 

The sign \pm is used because both the positive and negative square roots of 8 produce 8 when squared. Similarly:

{ (a-2 })^{ 2 }=7\\ \\ a-2=\pm \sqrt { 7 } \\ \\ a=\pm \sqrt { 7 } +2

This section will discuss two different methods to solve quadratic equations, while another method is discussed in the Quadratic Formula section.

 

Solving by Factorisation

The first method require trinomial factorisation.

 

Example:

 

1. Solve  { x }^{ 2 }+2x-3=0

Show Answer

{ x }^{ 2 }+2x-3=0\\ \\ (x+3)(x-1)=0\\ \\ x+3=0\qquad or\qquad x-1=0\\ \\ x=-3\qquad or\qquad x=1

 

As seen above, the solutions could only be obtained if either x+3=0 or x-1=0. Therefore there are two possible solutions for this quadratic equations.

 

Remember: To be certain that a solution is correct, substitute the solution into the original equation and see if it is satisfied.

 

 

2. Solve  { x }^{ 2 }+6x+9=0

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{ x }^{ 2 }+6x+9=0\\ \\ (x+3)(x+3)=0\\ \\ { (x+3) }^{ 2 }=0\\ \\ x+3=\sqrt { 0 } =0\\ \\ x=-3

Since both solutions are equal, then we only have one solution x=-3

 

 

3. Solve  3{ a }^{ 2 }-14a=-8

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3{ a }^{ 2 }-14a=-8\\ \\ 3{ a }^{ 2 }-14a+8=0\\ \\ (3a-2)(a-4)=0\\ \\ 3a-2=0\qquad or\qquad a-4=0\\ \\ a=\frac { 2 }{ 3 } \qquad or\qquad a=4

 

 

Solving by Completing the Square

Not all trinomials will factorise perfectly, so other methods such as completing the square needs to be used to solve quadratic equations. Starting from the basics, completing the square is basically re-writing a quadratic equation in another form:

 

a{x}^{2}+bx+c=0\qquad to \qquad a{ (x+d) }^{ 2 }+e=0

where             d=\frac { b }{ 2a } \qquad and\qquad e=c-\frac { { b }^{ 2 } }{ 4a }

Using a simple example, the equation {x}^{2}+bx  can be rearranged into a square, and by completing the square with {\frac { b }{ 2 }}^{2}.

In geometry, it looks like:

{x}^{2}+bx  \qquad \qquad \qquad + \qquad\qquad \qquad {\frac { b }{ 2 }}^{2} \qquad \qquad \qquad = \qquad \qquad \qquad{ \left( x+\frac { b }{ 2 }  \right)  }^{ 2 }

(courtesy of mathsisfun.com)

Examples:

 

Complete the square on {x}^{2}+6x.

{ x }^{ 2 }+6x\qquad <-\quad \frac { 6 }{ 2 } =3\\ \\ { x }^{ 2 }+6x+{ 3 }^{ 2 }\quad =\quad { (x+3) }^{ 2 }\\ \\ { x }^{ 2 }+6x+9\quad =\quad { (x+3) }^{ 2 }

 

More advanced quadratic equations require balancing.

For example, complete the square on { x }^{ 2 }+6x+7.

{ x }^{ 2 }+6x+7\qquad <-\quad \frac { 6 }{ 2 } =3\\ \\ =\quad ({ x }^{ 2 }+6x+{ 3 }^{ 2 })+(7-{ 3 }^{ 2 })\\ \\ =\quad { (x+3) }^{ 2 }-2

 

Now, to solve a quadratic equation in the form of a{x}^{2}+bx+c=0 by completing the square, there are steps to follow:

1. Divide all terms by a (the coefficient of {x}^{2}

2. Move the independent term \frac{c}{a} to the right side.

3. Complete the square on the left side, and balance this by adding \frac{b}{2a} to the right hand.

4. Take the square root on both sides.

5. Subtract the number that remains on the left side to finally obtain x.

 

Examples:

Solve by completing the square:

1. { x }^{ 2 }-6x+3=0

Show Answer

 { x }^{ 2 }-6x=-3\\ \\ { x }^{ 2 }-6x+9=-3+9\\ \\ { (x-3) }^{ 2 }=6\\ \\ x-3=\pm \sqrt { 6 } \\ \\ x=\pm \sqrt { 6 } +3

 

2. 5{ x }^{ 2 }-4x-2=0

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{ x }^{ 2 }-\frac { 4 }{ 5 } x-\frac { 2 }{ 5 } =0\\ \\ { x }^{ 2 }-\frac { 4 }{ 5 } x=\frac { 2 }{ 5 } \qquad <-\quad \cfrac { \frac { 4 }{ 5 }  }{ 2 } =\frac { 4 }{ 10 } \\ \\ { x }^{ 2 }-\frac { 4 }{ 5 } x+{ \left( \frac { 4 }{ 10 }  \right)  }^{ 2 }=\frac { 2 }{ 5 } +{ \left( \frac { 4 }{ 10 }  \right)  }^{ 2 }\\ \\ { \left( x-\frac { 4 }{ 10 }  \right)  }^{ 2 }=\frac { 14 }{ 25 } \\ \\ x-\frac { 4 }{ 10 } =\pm \sqrt { \frac { 14 }{ 25 }  } \\ \\ x=\pm \sqrt { \frac { 14 }{ 25 }  } +\frac { 4 }{ 10 }