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

quadratic equations

Quadratic equations are type of equations shown in this form:

standard form quadratic equations

Where a ≠ 0, b, c are given real numbers. Every x that satisfies that equation is called the solution of quadratic equation. If a = 0, our equation will be bx + c = 0 which transforms into a linear equation. This means that a must be different from zero.

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

There are three possible forms of a quadratic equations.

First form is the form in which \ b = 0 or \ ax^2 + c = 0. You solve it by switching the c to the other side, dividing by a and taking its square root.

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

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

In general, all quadratic equations will have exactly two solutions.
 

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

The procedure is the same, and soon enough we got to the point where \ x= \sqrt{(-4)}. This means that our equation has no real solutions, but only complex: \ x_1 = -2i, \ x_2 = 2i.

The second form is the form in which \ c = 0 or \ ax^2 + bx = 0. This is one of the simplest quadratic equations because they can be written as a product of two linear equations, where one solution is always zero.

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

First we’ll extract x. \ 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. First case is where x equals to zero, and second is where \ 5x + 4 equals to zero. This directly leads us to our solutions: \ x_1 = 0 and \ 5x + 4 = 0 => \ 5x = - 4 => \ x = - \frac{5}{4}.

The third form is the hardest form where every coefficient is different from zero. The universal way of solving these kinds of quadratic equations is using the formula:

quadratic equations formula

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

First we determine our coefficients: \ a = 1, \ b = 5 and \ c = 4. Now we simply insert it into our formula.

solving example 4

Of course, if we insert the solutions we got into our given equation we should get a zero.

\ (-1)^2 - 5 + 4 = 0
 

\ (-4)^2 - 20 + 4 = 16 - 20 + 4 = 0
 

This is the universal way of finding solutions but 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 second form we showed.

Example 5.

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

\ x^2 + 6x + 9 = (x+3)^2 => \ (x+3)^2 = 0 => \ x + 3 = 0
 

This means that our solutions is \ x = -3. But we got only one solution, and we said that every quadratic equation has two solutions. Did we do something wrong? Let’s try to solve it with formula then. \ a = 1, \ b = 6, \ c = 9.

We got only one solution again. This happens when we can transform our whole term in equation with a whole square. In this case we can write:

\ (x+3)^2 = 0 as \ (x+3)(x+3) = 0.
 

We say that -3 is the double solution of equation \ (x+3)^2 = 0.

Nature of solutions will always depend on the term underneath the square root in our formula for solutions of quadratic equation. This number is called the discriminant and is usually marked with D.

discriminant in quadratic equations

If \ D > 0 the equation has two real solutions, if \ D = 0 the equation will have two complex solutions, and if \ D = 0 the equation will have one double solution.

Another way of solving quadratic equations is completing the square. First, what does it mean to complete the square?

Example 6. If you have a term \ 4x^2 + 4x + 5 you can write it as \ (2x)^2 + 2*2x + 1 + 4. You have a first member squared, second doubled and third squared. You can write that down as \ (2x+1)^2 + 4.

What is the easiest way to learn how to complete the square?

complete the square

First you focus only on first two members, you’ll think of a linear coefficient later. The leading coefficient must be a whole square and if he is not you can make him become a whole square.

\ 8x^2 + 6x + 2
 

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

\ 2((2x))^2 + 6x) + 2
 

Now you have your first member squared. \ (a + b)^2 = a^2 + 2ab + b^2. We have to adjust that middle member so that we get something double.

\ 2((2x))^ 2 + 2*3x) + 2
 

Now we got a = 2x and b = 3 and we only have to handle with the free coefficient. 32 = 9 which means that the free coefficient we get when expanding our square is equal to 9. But we have to multiply it with 2 because whole square is multiplied with two. Our solution is:

\ 2(2x + 3)^2 + 2 - 18

\ 2(2x + 3)^2 - 16

How can this help us solve quadratic equations?

Example 7. Solve the equation \ 8x^2 + 6x + 2 = 0 by completing the square.

\ 8x^2 + 6x + 2 = 0

\ 2(2x + 3)^2 - 16 = 0

\ 2(2x + 3)^2 = 16

\ (2x+3)^2 = 8

Now we can take square root of whole equation:

\ 2x + 3 = ±\sqrt{8}

\ 2x = -3 ± \sqrt{8}

\ x = (-3 ± \sqrt{8})/2

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.

If x1 and x2 are the solutions of quadratic equation \ ax^2 + bx + c = 0, then they are also solutions to

vietas formulas

These formulas are called Vieta’s formulas.

Example 7. If \ x = 3 is one solution to the equation \ x^2 + bx + 8 = 0, what is the other solution and what is b?

According to the Vieta’s formulas we get that:

\ x_1 + x_2 = -\frac{b}{1} and \ x_1x_2 = \frac{8}{1}. Since we already have one solution we can write this as:

\ 2 + x_2 = -b, \ 2x_2 = 8. From the second equation we get that \ x_2 = 4. Which leads us to \ b = -6.

Example 8. Determine the product and sum of solutions of equation \ x^2 + 2x - 5 = 0.

\ a = 1, \ b = 2, \ c = -5 => \ x_1 + x_2 = -2 , \ x_1x_2 = -5

 

System of linear and quadratic equation

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 can’t be solved without any terms attached to it, but if we have one more additional condition like linear equation it can be solved. This is because with that linear equation we get the information in which relation are x and y and we can extract one using the other to get quadratic equation with only one unknown. The solutions to this system are two ordered number pairs (x1, y1) and (x2, y2).

Example 9. Solve the system of linear and quadratic equation

\ x^2 + 2xy - 6y^2 = 0

\ x + y = 2

First we’ll observe linear equation and extract one unknown according to the other one. It doesn’t matter which unknown you choose.

\ y = 2 - x
 

Now you take that y and insert him into a 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

Biquadratic equations

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

Example 10. Solve the following system

\ xy = 1

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

First we’d like to have only one unknown in our equation so we’ll extract y = 1/x and insert it in other equation.

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

\ x^4 - 4 + 3x^2 = 0

\ x^4 + 3x^2 - 4 = 0

Now we can make a substitution. We’d like to get rid of x to the power of four because we still don’t know how to calculate with those powers. Also, we should be able to turn this equation to a quadratic equation. We can substitute \ x^2 with u and get:

\ u^2 + 3u - 4 = 0

\ u1 = 1, \ u2 = -4

We are not done yet, we have to get back to our substitution and solve these smaller quadratic equations.

\ x^2 = u
 

For u_1 we have: \ x^2 = 1 => \ x1 = 1, \ x2 = -1

For u_2 we have: \ x^2 = -4 => \ x3 = -2i, \ x4 = 2i

Quadratic equations worksheets

  Solve by taking square roots (222.8 KiB, 206 hits)

  Solve by factoring (466.1 KiB, 264 hits)

  Completing the square (345.0 KiB, 249 hits)

  Solve using quadratic formula (308.2 KiB, 243 hits)

  Find a discriminant (473.2 KiB, 159 hits)

  Factoring quadratic expressions (315.0 KiB, 178 hits)

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