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'''One root equations'''
 
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'''Written by Justin Stevens '''
 
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==  Deriving the quadratic formula ==
 
==  Deriving the quadratic formula ==
  
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When <math>4a=b=c</math> we will have one root.  That one root will be –2.
 
When <math>4a=b=c</math> we will have one root.  That one root will be –2.
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== Contributors ==
 
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Thanks to our contributors:
 
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AIME15
 
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El_Ectric
 
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BOGTRO
 
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bzprules
 
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El_Ectric
 
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SuperNerd123
 

Latest revision as of 18:50, 2 June 2011

Deriving the quadratic formula

When we have a quadratic, we normally specify it in the form $ax^2+bx+c=0$

The way to find a formula for the roots of the equations goes as follows:

\[ax^2+bx+c=0\] \[\frac{ax^2+bx+c}{a}=\frac{0}{a}\] \[x^2+\frac{bx}{a}+\frac{c}{a}=0\] \[\text{We complete the square to get}\] \[(x+\frac{b}{2a})^2+\frac{c}{a}=\frac{b^2}{4a^2}\] \[(x+\frac{b}{2a})^2=\frac{b^2}{4a^2}-\frac{c}{a}\] \[(x+\frac{b}{2a})^2=\frac{b^2}{4a^2}-\frac{c}{a}*\frac{4a}{4a}\] \[(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}\] 

\[\sqrt{(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}\] (Error compiling LaTeX. Unknown error_msg)

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

We have two real roots when $\sqrt{b^2-4ac}> 0$ We have complex roots when $\sqrt{b^2-4ac}<0$ We have one root when $\sqrt{b^2-4ac}=0$

We are going to analyze the case when $\sqrt{b^2-4ac}=0$

$b^2-4ac$ is called the discriminant.

Square both sides to get: $b^2-4ac=0$ Add $4ac$ to both sides to get:

$b^2=4ac$


Let’s evaluate cases now.

The discriminant when a= c

First we evaluate the case $a=c$

When a=c, we have:

$b^2=4a^2$

$b^2=4c^2$

Take the square root of both sides to get: $b=2c$ $b=2a$

Therefore when $a=c$ and $b=2a$ we are going to have one root for the equation.

Example 1

$9x^2+18x+9x=0$

$\frac{-18 \pm \sqrt{(18)^2-4*9*9}}{2*9}$

$\frac{-18 \pm \sqrt{(9^2*2^2)-(9^2*2^2)}}{2*9}$

$\frac{-18 \pm \sqrt{0}}{18}$

$\frac{-18}{18}=\boxed{-1}$


Example 2

$x^2+2x+1=0$

$\frac{-2 \pm \sqrt{2^2-4*1*1}}{2*1}$

$\frac{-2 \pm \sqrt{4-4}}{2}$

$\frac{-2 \pm \sqrt{0}}{2}$

$\frac{-2}{2}=\boxed{-1}$

When $a=c$ and $b=2a$ we can write the problem as follows:

$ax^2+2ax+a=0$

Using the quadratic formula, we get: $\frac{-2a \pm \sqrt{(2a)^2-4*a*a}}{2*a}$

$\frac{-2a \pm \sqrt{4a^2-4a^2}}{2a}$

$\frac{-2a \pm \sqrt{0}}{2a}$

$\frac{-2a}{2a}=\boxed{-1}$

A simpler way to do this, is to start off by dividing by a. We get: $x^2+2x+1=0$ $(x+1)^2=0$ $x=\boxed{-1}$

If we want to write this as one equality, we can do as follows:

$a=c$ $b=2a$

Divide by 2 on the second equation to get:

$\frac{b}{2}=a$

Therefore we have $a=c=\frac{b}{2}$

So, when $a=c=\frac{b}{2}$ we will have one root, and that one root will be –1.

The discriminant when a=b

When $a=b$ we get:

$b^2-4ac=0$ $a^2-4ac=0$ $a(a-4c)=0$

The roots are going to be $a=0, a=4c$

Therefore we have: $a=b=0$ or $a=b=4c$

However, if (a,b)=0, we would be left with $(0*x^2+0x+c)=0$ This is only a constant term left.

So, our only case is $a=b=4c$

Example 1

$4x^2+4x+1=0$

$\frac{-4 \pm \sqrt{4^2-4*4*1}}{2*4}$

$\frac{-4 \pm \sqrt{16-16}}{8}$

$\frac{-4}{8}=\boxed{\frac{-1}{2}}$

Example 2

$8x^2+8x+2=0$

$\frac{-8 \pm \sqrt{8^2-4*8*2}}{2*8}$

$\frac{-8 \pm \sqrt{64-64}}{16}$

$\frac{-8}{16}=\boxed{\frac{-1}2}$

Let’s find a general formula for this. If $a=b=4c$, we have:

$4cx^2+4cx+c=0$

Using the quadratic formula, we get: $\frac{-4c \pm \sqrt{4c^2-4*4c*c}}{2*4c}$

$\frac{-4c \pm \sqrt{4c^2-4c^2}}{8c}$

$\frac{-4c \pm 0}{8c}$

$\frac{-4c}{8c}$

$\frac{-4*\cancel{c}}{8*\cancel{c}}=\boxed{\frac{-1}2}$

Another way to do this, is to start off by dividing by c. $4x^2+4x+1=0$ Divide by 4 to get: $x^2+x+\frac{1}{4}$

$(x+\frac{1}{2})^2=0$

$x=\boxed{\frac{-1}{2}}$

Therefore when $a=b=4c$ we will have one root which will be $\frac{-1}{2}$

The discriminant when b=c

$b^2-4ac=0$ $b=c$, therefore $c^2-4ac=0$ $c(c-4a)=0$

The roots are going to be $c=0$ which gives us: $0(0-4a)=0$

Since $b=c$ we get:

$ax^2+0*x+0=0$ $ax^2=0$

Divide by a to get: $x^2=0$ $x=0$

However, since (a,b,x)=(0,0,0) this is just a case with only a quadratic term, and shouldn’t be considered.

Let’s try the second case of $c(c-4a)$ which is going to be $c=4a$

We have: $b=c=4a$


Example 1

$x^2+4x+4=0$

$(x+2)^2=0$

$x=-2$

Example 2

$2x^2+8x+8=0$

$\frac{-8 \pm \sqrt{8^2-4*8*2}}{2*2}$

$\frac{-8 \pm 0}{4}$

$\frac{-8}{4}=-2$

Let’s try an in-general example of when $b=c=4a$

$ax^2+4ax+4a=0$

Using the quadratic formula, we get $x=\frac{-4a \pm \sqrt{(4a)^2-4*4a*a}}{2*a}$

$x=\frac{-4a \pm \sqrt{16a^2-16a^2}}{2a}$

$x=\frac{-4a}{2a}=-2$

We can also start off by dividing by a to get: $x^2+4x+4=0$ $(x+2)^2=0$ $x=\boxed{-2}$

Therefore when $b=c=4a$ in $ax^2+bx+c=0$, the root is going to be $-2$

Conclusion

When $a=\frac{b}{2}=c$ we will have one root. That one root will be –1.

When $a=b=4c$ we will have one root. That one root will be $\frac{-1}{2}$.

When $4a=b=c$ we will have one root. That one root will be –2.

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