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Form of quadratic equations, discriminant formula, Vieta’s formulas, biquadratic equations

quadratic equations

The quadratic equation

Quadratic equations are the type of equations shown in the form below:

    \[ax^2 + bx + c = 0,\]

where a \neq 0,  b,  c are  real numbers. Every x (real or complex) that satisfies this equation is called the solution of the quadratic equation.

If a = 0, our equation will be equal to bx + c = 0 which transforms into a linear equation. This means that a must be different from zero.

Number a in this equation is called the leading coefficient, number b is the linear coefficient and c free coefficient.

There are three special forms of  quadratic equations:

1.)  If \ b = 0 and \ c \neq 0 then the equation is:

    \[ax^2 + c = 0.\]

First, from the left and right sides of an equal sign we subtract c and then divide the entire equation by a. We now get an equivalent equation:

    \[x^2 = - \frac {c}{a}.\]

This equation always has two different solutions. The solutions may be real or complex numbers, depending on the sign of numbers a and c.

If n= - \frac {c}{a} then the quadratic equation x^2 = n , n \neq 0, has two solutions:

a) If n > 0, then the solutions are real numbers x_1 = \sqrt{n} and x_2 =-\sqrt{n}.

b) If n < 0, then the solutions are complex numbers x_1 = i \sqrt{|n|} and x_2 = - i \sqrt{|n|}.

Example 1. Solve the equation \ 2x^2 - 8 = 0.

Solution:

\ 2x^2 = 8 \Rightarrow  x^2 = 4 and we got \ x = \sqrt{4}. As we know \ 2^2 = 4, but also \ (-2)^2 = 4 which means that our solutions are \ x_1 = 2 and \ x_2 = -2.

 Example 2. Solve the equation \ 2x^2 + 8 = 0.

Solution:

The procedure is the same, and quickly enough we arrived at the point where \ x= \sqrt{(-4)}. This means that  solutions of our quadratic equation are not real numbers, and are only complex: \ x_1 = -2i, \ x_2 = 2i.

2.) \ c = 0 and b \neq 0.

The quadratic equation is now given in the form:

    \[\ ax^2 + bx = 0.\]

This is one of the simplest forms of a quadratic equation because it can be written as a product of two linear equations, where one solution is always equal to x_1 = 0 and other is equal to x_2 = -\frac{b}{a}.

Example 3. Solve the equation \ 5x^2 + 4x = 0.

Solution:

First, we will extract x. Then we have  \ x(5x + 4) = 0. Now we have a product of two numbers that is equal to 0. This means that one of them has to be zero. The first case is where x equals to zero, and second is where \ 5x + 4 equals to zero. This directly leads us to our two solutions which are \ x_1 = 0 and \ 5x + 4 = 0 => \ 5x = - 4 => \ x = - \frac{5}{4}.

3.) b = 0 and c = 0.

In this case the quadratic equation looks like:

    \[ax^2 = 0.\]

This equation is valid iff is x=0. We will say that this quadratic equation also has two solutions which are equal, that is x_1 = x_2 = 0.

x = 0 is called a double solution of the quadratic equation ax^2 = 0.

 

The solutions of the quadratic equation

 

We observe the quadratic equation:

    \[ax^2 + bx + c = 0,  \quad a\neq 0, \quad b, c \in \mathbb{R}\]

The universal method of solving these kinds of quadratic equations is by using the formula for the solutions of the quadratic equation:

    \[x_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\]

 

Example 4. Solve the equation \ x^2 + 5x + 4 = 0.

Solution:

First we determine our coefficients: \ a = 1, \ b = 5 and \ c = 4. Now we simply inserted it into the formula for the solutions of the quadratic equation:

 

    \[x_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac {-5 \pm \sqrt{5^2 - 4\cdot\ 1 \cdot 4}}{2 \cdot 1} = \frac{-5 \pm 3}{2}\]

    \[\Rightarrow x_1 = \frac{-5 + 3}{2} = -1 \quad  and  \quad  x_2 = \frac{-5 - 3}{2} = -4.\]

 

If we insert the solutions we obtained into the given equation, left side of the equation must be equal to 0.

This is the universal method of finding solutions, however, in certain cases we can make it a lot easier. One way of simplifying is factoring. If we have two factors whose product is equal to zero we can easily find solutions just like in the third special form of the quadratic equation we observed.

Example 5.

Solve the equation \ x^2 + 6x + 9 = 0 by factoring.

Solution:

    \[x^2 + 6x + 9 = (x+3)^2  \Rightarrow  (x+3)^2 = 0 .\]

We can write:

    \[(x+3)^2 = 0  \Leftrightarrow  (x+3)(x+3) = 0.\]

 

So, -3 is the double solution of an equation \ (x+3)^2 = 0.

 

Solving quadratic equations  by completing the square

 

Recall the formula for the square of a binomial:

    \[(a+b)^2 = a^2 + 2ab + b^2\]

to proceed further.

Example 6.

If we have a term such as \ 4x^2 + 4x + 5 we can express it as \ (2x)^2 + 2\cdot 2x + 1 + 4.

We have a first member squared, second doubled and third squared.

We can write that down as \ (2x+1)^2 + 4.

 

complete the square

 

Example 7.

We are observing the quadratic equation 8x^2 + 6x  + 2 = 0.

Let’s write the equation now in form:

\ 8x^2 + 6x  = -2.
 

From the left and right side of the equation we previously subtracted -2. We apply a methodology of completing the square on the left side of the equation.

8 is not a whole square, but 4 is. This means that we can first extract 2 and get:

 

\ 2\cdot (2x)^2 + 6x = -2.
 

Now we have the first member squared. It can be observed that the square of the second member is missing, which is 3, because the first member when multiplied by 3 is equal to 6x. Also, we need to multiply 6x by 2 because the first member was multiplied by 2.

Therefore, now we have:

    \[2 \cdot (2x)^ 2 + 2\cdot 3x = -2 .\]

 In the next step, the square from the second member, which is equal to 3^2 = 9, we added to both sides of the equation also multiplied by two:

    \[2 \cdot (2x)^2 + 2 \cdot 3x + 2 \cdot 3^2 = -2 +2\cdot 3^2\]

    \[2 \cdot (2x + 3)^2 = -2 + 18\]

    \[2 \cdot(2x + 3)^2 = 16\]

    \[(2x + 3)^2 = 8.\]

How can this help us solve quadratic equations?

Example 8. Solve the equation \ 8x^2 + 6x + 2 = 0 by completing the square.

Solution:

From the previous example above it was expressed that the quadratic equation \ 8x^2 + 6x + 2 = 0 can be written as:

    \[(2x + 3)^2 = 8.\]

Now we can calculate the square root of the whole equation:

\ 2x + 3 = \pm \sqrt{8}

\ 2x = -3 \pm \sqrt{8}

\ x = \frac{-3 \pm \sqrt{8}}{2}.

The discriminant of the quadratic equation

 

 The discriminant of the quadratic equation ax^2 + bx + c is the number D:

    \[D=b^2 - 4ac.\]

Then the solutions of the quadratic equation written by using the discriminant are:

    \[x_{1,2}= \frac{-b \pm \sqrt{D}}{2a}.\]

If \ D > 0, then the quadratic equation thus has two distinct real solutions, if \ D < 0, the quadratic equation has two solutions that are complex conjugates , and if \ D = 0 the quadratic equation has one real solution of multiplicity two.

 

Vieta’s formulas

 

French mathematician François Viète also studied quadratic equation and came to an important relation between quadratic equation and system of two equations with two unknowns. Those unknowns are the solutions to observed quadratic equation.

The solutions x_1 and x_2  of the quadratic equation \ ax^2 + bx + c = 0 meet the criteria of the Vieta’s formulas:

    \[x_1 + x_2 = -\frac{b}{a},\]

    \[x_1 \cdot  x_2 = \frac{c}{a}.\]

 

Example 9.

If \ x = 3 is one solution to the equation \ x^2 + bx + 8 = 0, what is the other solution and what is b?

Solution:

According to the Vieta’s formulas we find that:

    \[x_1 + x_2 = -\frac{b}{1}\]

    \[x_1 \cdot x_2 = \frac{8}{1}.\]

Since we already have one solution this can be written as:

    \[2 + x_2 = -b\]

    \[2x_2 = 8.\]

From the second equation we find that \ x_2 = 4, which leads us to \ b = -6.

Example 10.

Determine the product and sum of the solutions of equation \ x^2 + 2x - 5 = 0.

Solution:

From the equation we read coefficients:

    \[a = 1,  b = 2, c = -5\]

.

Now, by using Vieta’s formulas we have:

    \[x_1 + x_2 = - \frac{b}{a}= -2, \quad  x_1\cdot x_2 = \frac{c}{a} = -5\]

.

The system of linear and quadratic equations

The quadratic equation with two unknowns has the form:

    \[Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0,\]

 

where A, B, C, D, E, F are real numbers – coefficients.

This kind of equation cannot be solved without any terms attached to it, however, if we have one more additional condition like a linear equation, it can be solved. This is because from the linear equation we get the information in which relation are unknowns x and y and we can then extract one using the other to get the quadratic equation with only one unknown. The solutions to this system are two ordered number pairs (x_1, y_1) and (x_2, y_2).

Example 11.

Solve the following system of linear and quadratic equation:

    \[x^2 + 2xy - 6y^2 = 0\]

    \[x + y = 2.\]

Solution:

First we will observe a linear equation and extract one unknown according to the other one. It doesn’t matter which unknown we choose. In this case, we choose y:

    \[y = 2 - x.\]

Now we take y and insert it into the quadratic equation to get quadratic equation with only one unknown:

\ x^2 + 2x(2-x) - 6(2-x)^2 = 0

\ x^2 + 4x - 2x^2 - 6(4 - 4x + x^2) = 0

\ -x^2 + 4x - 24 + 24x - 6x^2 = 0

\ -7x^2 + 28x - 24 = 0

\ 7x^2 - 28x + 24 = 0.

 

The solutions of the quadratic equation \7x^2 - 28x +24 = 0 are:

    \[x_{1,2} = \frac{28 \pm \sqrt{28^2 - 4 \cdot 7 \cdot 24}}{2\cdot 7} = \frac{28 \pm \sqrt{784-672}}{14} =\frac{28 \pm \ 4\sqrt{7}}{14} =  2 \pm \frac{2\sqrt{7}}{7}\]

.

Now we must solve linear equations y_1 = 2 - x_1  and  y_2 = 2 - x_2.

As follows:

    \[y_1 = 2 - 2 - \frac{2\sqrt{7}}{7} = - \frac{2\sqrt{7}}{7},\]

    \[y_2 = 2 - 2 + \frac{2\sqrt{7}}{7} =   \frac{2\sqrt{7}}{7}.\]

 

The solutions are: (2+\frac{2\sqrt{7}}{7}, - \frac{2\sqrt{7}}{7}), and (2-\frac{2\sqrt{7}}{7}, \frac{2\sqrt{7}}{7}).

 

Biquadratic equations

 

Biquadratic equations are equations with two unknowns to the power of two. They are solved by using substitution.

Example 12. Solve the following system:

\ xy = 1

\ x^2 - 4y^2 = -3.

 

First we would like to have only one unknown in our equation so we will extract by y =\frac{1}{x} and insert it in other equation:

    \[x^2 - 4 \frac{1}{x^2} = -3\]

    \[x^2 - \frac{4}{x^2 + 3} = 0 / \cdotx^2\]

    \[x^4 - 4 + 3x^2 = 0\]

    \[x^4 + 3x^2 - 4 = 0.\]

Now we can make a substitution. We will remove of x to the power of 4 through substitution \ x^2 with u to get the quadratic equation:

    \[u^2 + 3u - 4 = 0\]

    \[\Rightarrow \ u_1 = 1 , u_2 = -4\]

.

We are still not finished, we have to return to our substitution and solve the quadratic equation:

\ x^2 = u.
 

For u_1 we have:

    \[x^2 = 1 \Rightarrow  x_1 = 1, \quad   x_2 = -1\]

.

For u_2 we have:

    \[x^2 = -4 \Rightarrow  x_3 = -2i, \quad  x_4 = 2i\]

.

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