Quadratic Equations

Quadratic Equations

A quadratic equation is written in the following form:

$ax^2+bx+c=0$

These equations contain polynomials of the second degree, with the square being the highest exponent.

It is important that we find the values of the constants $a$ , $b$ and $c$ . The standard form must be written in the above format.

The exponent of 2 must be present in order for it to be a quadratic equation. When the equation is given in another form, we can always re-arrange it to get $a>0$

Example: $12x-x^2=16$ We need to put this in the standard form:

$-x^2+12x=16$

$-x^2+12x-16=0$

Now multiply all by $-1$ to get the standard form: $x^2-12x+16=0$

Please note that $b$ and $c$ can be $0$ , $<0$ or $>0$ .

1.Solving by Factoring:

We have used factoring in our examples in this site. Now we can see how we arrived at these factors.

Given :

$ax^2+bx+c=0$

Solve by factoring.

(1) $\begin{equation*} \begin{split} ax^2+bx+c&=a(x^2+\frac{b}{a}x+\frac{c}{a})\\ &=a\left[{(x+\frac{b}{2a})}^2-{(\frac{b}{2a})}^2+\frac{c}{a}\right]\\ &=a\left[{(x+\frac{b}{2a})}^2-\frac{b^2}{4a^2}+\frac{4ac}{4a^2}\right]\\ &=a\left[{(x+\frac{b}{2a})}^2-\frac{(b^2-4ac)}{4a^2}\right]\\ &=a\left[{(x+\frac{b}{2a})}^2-{(\frac{\sqrt{b^2-4ac}}{2a})}^2\right] \end{split} \end{equation*}$

Now we have the difference of squares seen in the identities lesson:

(2) $\begin{equation*} \begin{split} a\left[{\left(x+\frac{b}{2a}\right)}^2-{\left(\frac{\sqrt{b^2-4ac}}{2a}\right)}^2\right]&=a\left[\left(x+\frac{b}{2a}+\frac{\sqrt{b^2-4ac}}{2a}\right)\left(x+\frac{b}{2a}-\frac{\sqrt{b^2-4ac}}{2a}\right)\right]\\ &=a\left[\left(x+\frac{b+\sqrt{b^2-4ac}}{2a}\right)\left(x+\frac{b-\sqrt{b^2-4ac}}{2a}\right)\right] \end{split} \end{equation*}$

Going back to our equation:

$ax^2+bx+c=a\left[\left(x+\frac{b+\sqrt{b^2-4ac}}{2a}\right)\left(x+\frac{b-\sqrt{b^2-4ac}}{2a}\right)\right]$

Finally we get:

From $ax^2+bx+c=0$

$a\left(x+\frac{b+\sqrt{b^2-4ac}}{2a}\right)\left(x+\frac{b-\sqrt{b^2-4ac}}{2a}\right)=0$

Any of the two factors will define one solution.

The two values of $x$ are:

$x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}$

$x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$

Solved by factoring.

2. Further Factoring methods

It may so happen that the roots are simple integers easy to find. We use the following method for factoring.

If we have $ax^2+bx+c$ We can try to find two values such:

-Their sum is $b$ -Their product is $ac$ Then we rewrite our expression by replacing $b$ by these two numbers. Example:

Factor:

$5x^2-21x-20$

Here

$a=5$

$b=-21$

$c=-20$

Find 2 numbers with a sum of $-21$ and a product of $(5)\cdot (-20)=-100$ .

With experience, we see that $-25$ and $4$ are the two numbers.

Now we write:

(3) $\begin{equation*} \begin{split} 5x^2-21x-20&=5x^2-25x+4x-20\\&=(5x^2-25x)+(4x-20)\\ &=\left[5x(x-5)+4(x-5)\right]\\&=(x-5)(5x+4) \end{split} \end{equation*}$

Finally: $5x^2-21x-20=(x-5)(5x+4)$

3- Using the quadratic formula

To solve $ax^2+bx+c=0$ , we are not going to factor every time. We use the formula we discovered earlier.

let’s use a discriminant $\Delta$ such: $\Delta=b^2-4ac$

From part 1 we get the following case:

-When $\Delta>0$ we have two roots for solution

$x=\frac{-b \pm \sqrt{\Delta}}{2a}$

-When $\Delta=0$ we have a duplicate root for solution

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

–When $\Delta<0$ there is no root in the real number system

(Since we do not have a square root for a negative number).

4. Completing the square:

We have nearly used it in part 1. We have to isolate the $x^2$ by moving everything else to the right side.

$ax^2+bx+c=0$

$x^2+\frac{b}{a}x=-\frac{c}{a}$

But we know from identities that:

$x^2+\frac{b}{a}x={(x+\frac{b}{2a})}^2-{(\frac{b}{2a})}^2$

So we simply get: ${(x+\frac{b}{2a})}^2-{(\frac{b}{2a})}^2=-\frac{c}{a}$

Then

${(x+\frac{b}{2a})}^2={(\frac{b}{2a})}^2-\frac{c}{a}$

${(x+\frac{b}{2a})}^2=\frac{b^2}{4a^2}-\frac{4ac}{4a^2}$

$(x+\frac{b}{2a})=\pm \sqrt{\frac{b^2}{4a^2}-\frac{4ac}{4a^2}}$

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

Finally, the solution with the two possible roots:

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

Example:Solve by completing the square:

$8x^2-3x=1116$

$x^2-\frac{3}{8}x=\frac{1116}{8}$

${(x-\frac{3}{16})}^2-{(\frac{3}{16})}^2=\frac{1116}{8}$

${(x-\frac{3}{16})}^2={(\frac{3}{16})}^2+\frac{1116}{8}$

${(x-\frac{3}{16})}^2=\frac{9}{256}+\frac{1116}{8}$

${(x-\frac{3}{16})}^2=\frac{9}{256}+\frac{35712}{256}$

${(x-\frac{3}{16})}^2=\frac{35721}{256}$

$x-\frac{3}{16}=\pm \sqrt{\frac{35721}{256}}$

$x=\frac{3}{16}\pm \frac{189}{16}$

$x_1=\frac{3+189}{16}$

$x_1=12$

$x_2=\frac{3-189}{16}$

$x_2=-\frac{93}{8}$

Answer: $(12, -\frac{93}{8})$

Relationships between the roots of a quadratic equation

When the equation has two roots $x_1$ and $x_2$ in $\mathbb{R}$ , we can note the following:

Sum of the roots

(4) $\begin{equation*} \begin{split} x_1+x_2&=\frac{-b+\sqrt{b^2-4ac}}{2a}+\frac{-b-\sqrt{b^2-4ac}}{2a}\\ &=\frac{-2b}{2a}\\ &=-\frac{b}{a} \end{split} \end{equation*}$

Finally:
$x_1+x_2=-\frac{b}{a}$

The product of the roots

(5) $\begin{equation*} \begin{split} x_1 \cdot x_2&=\left(\frac{-b+\sqrt{b^2-4ac}}{2a}\right) \cdot \left(\frac{-b-\sqrt{b^2-4ac}}{2a}\right)\\ &=\frac{c}{a} \end{split} \end{equation*}$

Finally:
$x_1 \cdot x_2=\frac{c}{a}$

Solve for $x$

$x^4-3x^2+1=0$

Please note that $x^4=(x^2)^2$

let $t=x^2$

We get:

$t^2-3t+1=0$

Here:

$a=1$

$b=-3$

$c=1$

(6) $\begin{equation*} \begin{split} \sqrt{\Delta}&=\sqrt{9-4\cdot 1 \cdot 1}\\ &=\sqrt{5} \end{split} \end{equation*}$

$t_1=\frac{3+\sqrt{5}}{2}$

Since $t=x^2$ we plug in:

$x^2=\frac{3+\sqrt{5}}{2}$

$x_1=\sqrt{\frac{3+\sqrt{5}}{2}}$

$x_2=-\sqrt{\frac{3+\sqrt{5}}{2}}$

$t_2=\frac{3-\sqrt{5}}{2}$

Since $t=x^2$ we plug in:

$x^2=\frac{3-\sqrt{5}}{2}$

$x_3=\sqrt{\frac{3-\sqrt{5}}{2}}$

$x_4=-\sqrt{\frac{3-\sqrt{5}}{2}}$

$Answer: (\sqrt{\frac{3+\sqrt{5}}{2}},-\sqrt{\frac{3+\sqrt{5}}{2}},\sqrt{\frac{3-\sqrt{5}}{2}},-\sqrt{\frac{3-\sqrt{5}}{2}})$

Solve for $x$ :

$8x^3-12x^2+2x-3=0$

We proceed by grouping:

$8x^3-12x^2+2x-3=0$

$8x^3+2x-12x^2-3=0$

$2x(4x^2+1)-3(4x^2+1)=0$

$(2x-3)(4x^2+1)=0$

Two cases:

$2x-3=0 \Leftrightarrow x=\frac{3}{2}$

This is the only solution:

Answer: $x=\frac{3}{2}$

Solve for $x$

$2x^{-2}=x^{-1}+1$

Let $x^{-1}=t$

We get:

$2t^2-t-1=0$

$a=2$

$b=-1$

$c=-1$

(7) $\begin{equation*} \begin{split} \sqrt{\Delta}&=\sqrt{(-1)^2-4 \cdot 2 \cdot (-1)}\\ &=\sqrt{9}\\&=3 \end{split} \end{equation*}$

$t_1=\frac{1+3}{4}=1$

But $t=x^{-1} \Rightarrow x^{-1}=1$

$x_1=1$

$t_2=\frac{1-3}{4}\=-\frac{1}{2}$

But $t=x^{-1} \Rightarrow x^{-1}=-\frac{1}{2}$

$x_2=-2$

Answer: $\{ -2, 1\}$

Solve for $x$

$\frac{3x}{x-4}=\frac{2x}{x-3}+\frac{6}{x^2-7x+12}$

But when we can factor: $x^2-7x+12=(x-3)(x-4)$

The Equation becomes: $\frac{3x}{x-4}=\frac{2x}{x-3}+\frac{6}{(x-3)(x-4)}$ with $x \not= 3, 4$

By using the common denominator: $3x(x-3)=2x(x-4)+6$

$3x^2-9=2x^2-8x+6$

$x^2-x-6=0$

$\sqrt{\Delta}=\sqrt{1+24}=\sqrt{25}=5$

The roots:

$x_1=\frac{1+5}{2}=3$

Not acceptable

$x_2=\frac{1-5}{2}=-2$

Answer: $\{-2\}$

Solve for $x$

$\sqrt{1+6x}=2-\sqrt{6x}$

Squaring both sides $1+6x=4-4\sqrt{6x}+6x$

$4\sqrt{6x}=3$

Squaring one more time $16(6x)=9$

$32x=3$

$x=\frac{3}{32}$

Verification shows $\frac{5}{4}$ on each side.

Answer: $\{\frac{3}{32} \}$

Quadratic functions

A quadratic function has the form
$f(x)=ax^2+bx+c$
As we have already seen, a,b and c are real numbers with $a \not=0$
The domain of the quadratic function is the set of all real numbers. $f(x)$ is defined for any value of $x$ .
$D=(-\infty,+\infty)$
Or simple:
$D=\mathbb{R}$ .
In the previous paragraphs, we solved the quadratic for $x$ . We can use the same concept here in graphing the quadratic functions.
x-intercepts:
These are the solutions found by solving $f(x)=0$ . We’ll see that this statement is the $x-axis$ . The value of $y$ is $0$ anywhere in this axis.
So we have 3 possibilities:
When $\Delta=\sqrt{b^2-4ac}$ is $>0$ , we had two values as roots. In graphing, the function crosses the $x-axis$ at these 2 points.

When $\Delta=\sqrt{b^2-4ac}$ is $=0$ , the function touches the $x-axis$ at that point.

When $\Delta=\sqrt{b^2-4ac}$ is $<0$ , the function never crosses or touches the $x-axis$ .

The $y-intercept$ is where the function crosses the $y-axis$ .

We can get it by making the variable $x$ equal to $0$ .

We need to re-write these functions in the following format

(8) $\begin{equation*} \begin{split} (ax^2+bx+c)&=\left[a \left(x+\frac{b}{2a}\right)^2-\left(\frac{b}{2a}\right)^2+\frac{c}{a}\right]\\ &=a\left(x+\frac{b}{2a}\right)^2+\frac{4ac-b^2}{4a} \end{split} \end{equation*}$

Let $h=-\frac{b}{2a}$
And
$k=\frac{4ac-b^2}{4a}$

The function can be written:

$f(x)=a{(x-h)}^2+k$

In graphing we’ll see that the point $(h,k)$ is the vertex of the function.
This is the point that defines the axis of symmetry.
$x=-\frac{b}{2a}$ is a vertical line that is the axis of symmetry of the function.
Example
Graph $f(x)=x^2-2x+1$
$h=-\frac{-2}{2}$
$h=1$
And $k=\frac{4\cdot1 \cdot1-2^2}{4}$
$k=0$

We re-write:
$f(x)=(x-1)^2$
The vertex is $(1,0)$
y-intercept: When $x=0$ , $f(x)=1$
x-intercept:
The solutions:
$\Delta=0$
It touches the $x-axis$ at the vertex.

Graphs variations

If we are using the form $f(x)=a(x-h)^2+k$ , we need to know how to graph these functions without long logics.

Let’s go from the base:

$f(x)=x^2$

Now for $f(x)=(x-1)^2$ , we just have to move this original graph 1 unit to the right.

Now let’s add to the y-coordinate

$f(x)=(x-1)^2+2$ , we just move the graph two units up.

Now if we change the sign of $a$

$f(x)=-x^2$ . The function is flipped about the $x-axis$ . It is the mirror of the original.

Now let’s make $a=2$ from the original

$f(x)=2x^2$

We can now say:

-If $a<0$ we flip the original before moving

-If $h>0$ , we move the original to the right by $h$ units.

-If $h<0$ , we move the original to the left by $h$ units.

-If $k>0$ , we move the original up by $k$ units.

-If $k<0$ , we move the original down by $k$ units.

-If $\left|a\right|>1$ the graph is narrower

-If $\left|a\right|<1$ the graph is wider.

Example:

$f(x)=-0.5(x+3)^2-2$

From $f(x)=x^2$ ,

flip, move 3 units to the left, move 2 units down. Graph is wider.

Grph the function:

$f(x)=x^2-9$

Problem 1

A certain equation has been graphed as follows. Please find the equation in the form of $f(x)=ax^2+bx+c$

Solution:

Click on the following to watch the video for the solution

Problem 2:

Given the following equation:

$mx+6-3mx^2=m+4x^2-3x$

1. What values of $m$ make this equation, an equation with 2 roots?

2. Find the values of $m$ so that the equation has only one root as a solution. Calculate the root in each case.

3. Find all the values of $m$ making the equation one without a solution in $\mathbb{R}$ .

4. Find the values of $m$ that give a solution of 4 as one of the roots.

5.Finally, find the values of $m$ making the product of the two roots $-10$ .

Solution:

Click here for the Solution in another post.

Problem 3:

Given the following equation:

$x^2+mx+9=0$

1. Find the positive value of $m$ so that the equation has only one root as a solution.

2.Calculate the root.

Solution:

$x^2+mx+9=0$
For a double solution, the discriminant $\Delta$ must be $=0$
$\Delta=m^2-4(1)(9)$
$m^2-36=0$
$m^2=36$
$m=6$
This is the only value per the prompt.
For $m=6$ we get:
$x^2+6x+9=0$
$x=\frac{-6}{2}$
$x=-3$