Difference between revisions of "Quadratic equation"

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(Expand page with details on methods of solving and examples)
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A '''quadratic equation''' can take two forms. The general formula, written as a function of <math>x</math>, <math>y=f(x)</math>, is:
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A '''quadratic equation''' is a [[polynomial]] of degree two.<ref name=britannica>{{cite web|url=https://www.britannica.com/science/quadratic-equation|title=Quadratic equation|accessdate=2019-07-29}}</ref> The general formula, written as a function of <math>x</math>, <math>y=f(x)</math>, is:
  
 
:<math>y = ax^2 + bx + c</math>
 
:<math>y = ax^2 + bx + c</math>
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:<math>ax^2 + bx + c = 0</math>
 
:<math>ax^2 + bx + c = 0</math>
  
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These are solved for using the [[quadratic formula]], which will not only solve for [[real number|real]] roots, but result in the [[imaginary number|imaginary]] roots if the parabola does not actually cross the y axis (this is when <math>4ac</math> is greater than <math>b^2</math>).
+
These are solved for using the [[quadratic formula]], which will not only solve for [[real number|real]] roots, but result in the [[imaginary number|imaginary]] roots if the parabola does not actually cross the x axis (this is when <math>4ac</math> is greater than <math>b^2</math>).
  
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Quadratic equations are very important in calculating the motion of bodies under constant [[acceleration]], i.e., [[gravity]] (when close to the earth's surface).
+
Quadratic equations are very important in calculating the motion of bodies under constant [[acceleration]], e.g., [[gravity]] (when close to the earth's surface).
  
 
The [[derivative]] of a quadratic equation is a simple [[linear function]]:
 
The [[derivative]] of a quadratic equation is a simple [[linear function]]:
  
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:<math>\frac{dy}{dx} = 2ax + b</math>
+
:<math>\frac{\mathrm{d}y}{\mathrm{d}x} = 2ax + b</math>
  
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== Factoring ==  
+
==Solving quadratic equations==
 +
There are several methods that can be used to solve quadratic equations. Different methods are easier to use depending on the equation one wishes to solve. Below are the main methods used.
  
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Quadratic equations can be simplified by factoring it into <math>(x + r)(x + s)</math>, where <math>r + s</math> equals B in <math>y = x^2 + bx + c</math> and <math>rs</math> equals <math>c</math> in the above equation.<ref>http://www.regentsprep.org/regents/math/algtrig/ate3/quadlesson2.htm</ref> Equations that cannot be easily factored this way can become easy to factor by [[completing the square]].  
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=== Factoring ===
 +
Solving by factorising is often the easiest for quadratics where ''a,b,c'' are all [[whole number]]s less than 20.<ref>{{cite web|url=https://www.mathsisfun.com/algebra/factoring-quadratics.html|title=Factoring Quadratics|accessdate=2019-07-29}}</ref> If this is not the case, one of the other methods will be more suitable.
  
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=== If the coefficient of X is not one ===
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Quadratic equations can always be written as the product of two linear factors of ''x'':
 +
:<math>
 +
ax^2 + bx + c = (x+r) \times (x+s)
 +
</math>
 +
When the equation is written in this form, it is said to be "factorized". Since we want to solve the quadratic equation being equal to zero, we can say that,
 +
:<math>
 +
(x+r) \times (x+s) = 0
 +
</math>
 +
which means that either (''x''+''r'') or (''x''+''s'') are equal to zero. This means that ''-r'' and ''-s'' are the solutions to the equation. To see how we can factorise the quadratic it's simplest to consider the case of ''a''=1 first,
 +
:<math>
 +
(x+r) \times (x+s) = x^2 + (r+s) x + r \times s = x^2 + bx + c
 +
</math>
 +
Comparing the coefficients of the ''x'' and constant terms, ''r'' and ''s'' should add together to equal ''b'' and multiply to give ''c''. To solve a quadratic equation by factorising, you should therefore try to guess a pair of numbers that meet these conditions. It is often easier to work methodically through pairs of [[factor]]s of ''c''. It is important to note that the solution will be the negative of these numbers.
  
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If the coefficient of X is not 1, then one can turn <math>ax^2 + bx</math> into <math>a(x^2 + b/ax)</math>. This works because both <math>x^2</math> and <math>bx</math> were divided by <math>a</math>, which was put back later via [[multiplication]].
+
To solve a quadratic equation when ''a'' is not 1, the ''a'' can simply be factored out,
 +
:<math>
 +
ax^2 + bx + c = a\left( x^2 + \frac{b}{a}x + \frac{c}{a} \right)
 +
</math>
 +
and the term in brackets solved as above.
 +
 
 +
====Example====
 +
Consider:
 +
:<math>
 +
x^2 + 5x + 6 = 0
 +
</math>
 +
We want to choose ''r'' and ''s'' so they add to give 5 and multiply to give 6. Suppose we pick ''r=1'' and ''s=6'' to begin with since these are factors of 6. We then have ''r×s=1×6=6'' so the multiplication condition is met. However 1+6=7 and not 5, so the addition condition is not satisfied; we have not found the solution.
 +
 
 +
Next suppose we try ''r=2'' and ''s=3''. Then we have ''r×s=2×3=6'' and ''r+s=2+3=5''. This satisfies both conditions so we can say the solution is (remembering the solutions are ''-r'' and ''-s'') ''x''=-2 and ''x''=-3.
 +
 
 +
===Completing the square===
 +
[[Completing the square]] is a particularly useful method that will always work, but takes longer and is more complicated than factorizing. The quadratic we want to solve is,
 +
:<math>
 +
ax^2 + bx + c = 0
 +
</math>
 +
Again consider the case of ''a''=1 for simplicity first. Completing the square works by first rewriting the equation into the form,
 +
:<math>
 +
x^2 + bx + c = \left(x + \alpha \right)^2 + \beta
 +
</math>
 +
where α and β are two numbers. Multiplying the right hand side of the equation out we see,
 +
:<math>
 +
x^2 + bx + c = x^2 + 2 \alpha x + \left(\alpha^2 + \beta \right)
 +
</math>
 +
We therefore have,
 +
:<math>
 +
\alpha = \frac{1}{2} b, \beta = c - \alpha^2
 +
</math>
 +
and the original quadratic can be rewritten as,
 +
:<math>
 +
x^2 + bx + c = \left(x + \frac{1}{2}b \right)^2 + c - \frac{1}{4}b^2
 +
</math>
 +
From this it is straightforward to solve and get both solutions. Taking the constant terms onto the right hand side,
 +
:<math>
 +
\left(x + \frac{1}{2}b \right)^2 = \frac{1}{4}b^2 - c
 +
</math>
 +
Taking the square root and remembering that we must take plus or minus,
 +
:<math>
 +
x + \frac{1}{2}b = \pm \frac{1}{2} \sqrt{b^2 - 4c}
 +
</math>
 +
Then subtracting the ''b''/2 from both sides gives the solution:
 +
:<math>
 +
x = -\frac{b \pm \sqrt{b^2 - 4c}}{2}
 +
</math>
 +
The case of ''a'' not equal to 1 can be solved by factoring it out and leads to the [[quadratic formula]].
 +
 
 +
====Example====
 +
Consider:
 +
:<math>
 +
x^2 + 3x + 2 = 0
 +
</math>
 +
We first write it in completed square form as,
 +
:<math>
 +
\left(x + \frac{3}{2} \right)^2 + 2 - \left(\frac{3}{2} \right)^2 = 0
 +
</math>
 +
Taking the right two terms onto the right hand side produces,
 +
:<math>
 +
\left(x + \frac{3}{2} \right)^2 = \frac{9}{4} - 2 = - \frac{1}{4}
 +
</math>
 +
Then square rooting and remembering the plus and minus sign,
 +
:<math>
 +
x + \frac{3}{2} = \pm \frac{1}{2}
 +
</math>
 +
Therefore the solutions are,
 +
:<math>
 +
x = \frac{3}{2} \pm \frac{1}{2}
 +
</math>
 +
which is ''x''=2 when we use the plus and ''x''=1 when we choose the minus.
 +
 
 +
===Quadratic formula===
 +
The [[quadratic formula]] is a [[formula]] that will always give the two solutions of a quadratic equation. It gives the solutions of a quadratic equation as,
 +
:<math>
 +
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
 +
</math>
 +
To use it simply substitute in for the different values of ''a, b'' and ''c''. If the quadratic equation you want to solve has unusual numbers or even algebraic expressions for any of ''a, b'' or ''c'', this is likely the fastest and easiest way to solve it.
  
 
== References ==
 
== References ==

Revision as of 12:33, July 29, 2019

A quadratic equation is a polynomial of degree two.[1] The general formula, written as a function of <math>x</math>, <math>y=f(x)</math>, is:

<math>y = ax^2 + bx + c</math>

The graph of a quadratic equation is a parabola, one of the conic sections. (In some cases, the parabola collapses, most obviously when <math>a = 0</math>)

The points where this curve crosses the x axis are represented by the second form of the equation:

<math>ax^2 + bx + c = 0</math>

These are solved for using the quadratic formula, which will not only solve for real roots, but result in the imaginary roots if the parabola does not actually cross the x axis (this is when <math>4ac</math> is greater than <math>b^2</math>).

Quadratic equations are very important in calculating the motion of bodies under constant acceleration, e.g., gravity (when close to the earth's surface).

The derivative of a quadratic equation is a simple linear function:

<math>\frac{\mathrm{d}y}{\mathrm{d}x} = 2ax + b</math>

Solving quadratic equations

There are several methods that can be used to solve quadratic equations. Different methods are easier to use depending on the equation one wishes to solve. Below are the main methods used.

Factoring

Solving by factorising is often the easiest for quadratics where a,b,c are all whole numbers less than 20.[2] If this is not the case, one of the other methods will be more suitable.

Quadratic equations can always be written as the product of two linear factors of x:

<math>

ax^2 + bx + c = (x+r) \times (x+s) </math> When the equation is written in this form, it is said to be "factorized". Since we want to solve the quadratic equation being equal to zero, we can say that,

<math>

(x+r) \times (x+s) = 0 </math> which means that either (x+r) or (x+s) are equal to zero. This means that -r and -s are the solutions to the equation. To see how we can factorise the quadratic it's simplest to consider the case of a=1 first,

<math>

(x+r) \times (x+s) = x^2 + (r+s) x + r \times s = x^2 + bx + c </math> Comparing the coefficients of the x and constant terms, r and s should add together to equal b and multiply to give c. To solve a quadratic equation by factorising, you should therefore try to guess a pair of numbers that meet these conditions. It is often easier to work methodically through pairs of factors of c. It is important to note that the solution will be the negative of these numbers.

To solve a quadratic equation when a is not 1, the a can simply be factored out,

<math>

ax^2 + bx + c = a\left( x^2 + \frac{b}{a}x + \frac{c}{a} \right) </math> and the term in brackets solved as above.

Example

Consider:

<math>

x^2 + 5x + 6 = 0 </math> We want to choose r and s so they add to give 5 and multiply to give 6. Suppose we pick r=1 and s=6 to begin with since these are factors of 6. We then have r×s=1×6=6 so the multiplication condition is met. However 1+6=7 and not 5, so the addition condition is not satisfied; we have not found the solution.

Next suppose we try r=2 and s=3. Then we have r×s=2×3=6 and r+s=2+3=5. This satisfies both conditions so we can say the solution is (remembering the solutions are -r and -s) x=-2 and x=-3.

Completing the square

Completing the square is a particularly useful method that will always work, but takes longer and is more complicated than factorizing. The quadratic we want to solve is,

<math>

ax^2 + bx + c = 0 </math> Again consider the case of a=1 for simplicity first. Completing the square works by first rewriting the equation into the form,

<math>

x^2 + bx + c = \left(x + \alpha \right)^2 + \beta </math> where α and β are two numbers. Multiplying the right hand side of the equation out we see,

<math>

x^2 + bx + c = x^2 + 2 \alpha x + \left(\alpha^2 + \beta \right) </math> We therefore have,

<math>

\alpha = \frac{1}{2} b, \beta = c - \alpha^2 </math> and the original quadratic can be rewritten as,

<math>

x^2 + bx + c = \left(x + \frac{1}{2}b \right)^2 + c - \frac{1}{4}b^2 </math> From this it is straightforward to solve and get both solutions. Taking the constant terms onto the right hand side,

<math>

\left(x + \frac{1}{2}b \right)^2 = \frac{1}{4}b^2 - c </math> Taking the square root and remembering that we must take plus or minus,

<math>

x + \frac{1}{2}b = \pm \frac{1}{2} \sqrt{b^2 - 4c} </math> Then subtracting the b/2 from both sides gives the solution:

<math>

x = -\frac{b \pm \sqrt{b^2 - 4c}}{2} </math> The case of a not equal to 1 can be solved by factoring it out and leads to the quadratic formula.

Example

Consider:

<math>

x^2 + 3x + 2 = 0 </math> We first write it in completed square form as,

<math>

\left(x + \frac{3}{2} \right)^2 + 2 - \left(\frac{3}{2} \right)^2 = 0 </math> Taking the right two terms onto the right hand side produces,

<math>

\left(x + \frac{3}{2} \right)^2 = \frac{9}{4} - 2 = - \frac{1}{4} </math> Then square rooting and remembering the plus and minus sign,

<math>

x + \frac{3}{2} = \pm \frac{1}{2} </math> Therefore the solutions are,

<math>

x = \frac{3}{2} \pm \frac{1}{2} </math> which is x=2 when we use the plus and x=1 when we choose the minus.

Quadratic formula

The quadratic formula is a formula that will always give the two solutions of a quadratic equation. It gives the solutions of a quadratic equation as,

<math>

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} </math> To use it simply substitute in for the different values of a, b and c. If the quadratic equation you want to solve has unusual numbers or even algebraic expressions for any of a, b or c, this is likely the fastest and easiest way to solve it.

References

  1. ↑ Quadratic equation. Retrieved on 2019-07-29.
  2. ↑ Factoring Quadratics. Retrieved on 2019-07-29.
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