Quadratic/Quartic Equation Definition

A second degree equation or, failing that, a quadratic one, with respect to an unknown, is expressed in the form:
\(a{x^2} + bx + c = 0\)
Where the unknown is \(x\), as long as \(a, b\) and c are real constants, with \(a \ne 0.\)

There are several techniques to solve quadratic equations, including factorization, in which case we must take into account the following property according to the resolution:

If the product of two numbers is zero then there are two possibilities:

1. Both are equal to zero.
2. If one is non-zero then the other is zero

The above can be expressed as follows:
If \(pq = 0\) then \(p = 0\) or \(q = 0\).

Practical example 1: solve the equation \({x^2} – 8\)=0

\({x^2} – 8 = 0\)	Initial situation
\({x^2} – 8 + 8 = 8\)	Add 8 to both sides of the equation to solve for \({x^2}\)
\(\sqrt {{x^2}} = \sqrt {{2^3}} = \sqrt {{2^2}2} = \sqrt {{2^2}} \sqrt 2 = 2\sqrt 2 \)	The square root is obtained looking for isolating \(x.\) 8 is factored and properties of radicals and powers are applied.
\(\left| x \right| = 2\sqrt 2 \)	You get the root of \({x^2}\)
\(x = \pm 2\sqrt 2 \)

The solutions of \({x^2} – 8\)=0 are:
\(x = – 2\sqrt 2 ,\;2\sqrt 2 \)

Practical example 2: Solve the equation \({x^2} – 144\)=0

\({x^2} – 144 = 0\)	Initial situation
\({x^2} – {12^2} = 0\)	The square root of 144 is 12. A difference of squares is identified.
\(\left( {x + 12} \right)\left( {x – 12} \right) = 0\)	The difference of squares is factored
\(x + 12 = 0\) \(x = – 12\)	We consider the possibility that the factor \(x + 12\) is equal to 0. The equation obtained is solved.
\(x – 12 = 0\) \(x = 12\)	We consider the possibility that the factor \(x – 12\) is equal to 0. The equation obtained is solved.

The solutions of the equation \({x^2} – 144 = 0\) are

\(x = – 12,\;12\)

Practical example 3: solve the equation \({x^2} + 3x = 0\)

\({x^2} + 3x = 0\)	Initial situation
\(x\left( {x + 3} \right) = 0\)	\(x\) is identified as a common factor and the factorization is performed.
\(x = 0\)	Consider the possibility that the factor \(x\) is equal to 0.
\(x + 3 = 0\) \(x = – 3\)	We consider the possibility that the factor \(x – 12\) is equal to 0. The equation obtained is solved.

The solutions of the equation \({x^2} + 3x = 0\), are:
\(x = – 3.0\)

Practical example 4: Solve the equation \({x^2} – 14x + 49 = 0\)

\({x^2} – 14x + 49 = 0\)	Initial situation
\({x^2} – 14x + {7^2} = 0\)	The square root of 49 is 7 and \(2x\left( 7 \right) = 14x.\) A perfect square trinomial is identified.
\({\left( {x – 7} \right)^2} = 0\)	The perfect square trinomial is expressed as a squared binomial.
\(x – 7 = 0\) \(x = 7\)

The solution of \({x^2} – 14x + 49 = 0\) is:
\(x = 7\)

Practical example 5: Solve the equation \(10{x^2} – 23x + 12 = 0\)

\(10{x^2} – 23x + 12 = 0\)	Initial situation
\(10{x^2} – 23x + 12 = 0\)	The product \(\left( {10} \right)\left( {12} \right) = 120 = \left( { – 8} \right)\left( { – 15} \right)\)
\(\left( {10{x^2} – 8x} \right) – 15x + 12 = 0\)	It is expressed as \( – 23x = – 18x – 15\)
\(2x\left( {5x – 4} \right) – 3\left( {5x – 4} \right) = 0\)	Identify \(2x\) as a common factor in the first addend and factor it. Identify \( – 3\) as a common factor in the second addend and factor it.
\(\left( {5x – 4} \right)\left( {2x – 3} \right) = 0\)	Factor the common factor \(5x – 4\)
\(5x – 4 = 0\) \(x = \frac{4}{5}\)	We consider the possibility that the factor \(5x – 12\) is equal to 0. The equation obtained is solved.
\(2x – 3 = 0\) \(x = \frac{3}{2}\)	Consider the possibility that the factor \(2x – 3\) is equal to 0. The equation obtained is solved.

The solutions of \(10{x^2} – 23x + 12 = 0\) are:
\(x = \frac{4}{5},\;\frac{3}{2}\)

Practical example 6: Solve the equation \({x^2} + 4x + 1 = 0\)

\({x^2} + 4x + 1 = 0\)	Initial situation The trinomial is not a perfect square
\({x^2} + 4x + 1 – 1 = – 1\)	Add -1 to each side of the equation.
\({x^2} + 4x = – 1\)	Since \(\frac{1}{2}\left( 4 \right) = 2\) by adding \({2^2}\), we get a perfect square.
\({x^2} + 4x + 4 = – 1 + 4\)	Add \({2^2}\;\) to each side of the equation. The left side is a perfect square.
\({\left( {x + 2} \right)^2} = 3\)	The perfect square trinomial is expressed as a squared binomial.
\(\sqrt {{{\left( {x + 2} \right)}^2}} = \pm \sqrt 3 \)	Take the square root of each side of the equation
\(\left| {x + 2} \right| = \sqrt 3 \) \(x = – 2 \pm \sqrt 3 \)	Solve for \(x\).

The solutions of \({x^2} + 4x + 1 = 0\) are:
\(x = – 2 – \sqrt 3 ,\; – 2 + \sqrt 3 \)

Practical example 7: Solve the equation \(5{x^2} + 3x – 1 = 0\)

\(5{x^2} + 3x – 1 = 0\)	Initial situation The trinomial is not a perfect square.
\(5{x^2} + 3x – 1 + 1 = 1\)	Add 1 to each side of the equation
\(\frac{1}{5}\left( {5{x^2} + 3x} \right) = \frac{1}{5}\left( 1 \right)\)	Multiply by each side of the equation so that the coefficient of \({x^2}\) equals 1.
\({x^2} + \frac{3}{5}x = \frac{1}{5}\)	product is distributed Since \(\frac{1}{2}\left( {\frac{3}{5}} \right) = \frac{3}{{10}}\), by adding \({\left( { \frac{3}{{10}}} \right)^2} = \frac{9}{{100}}\) gives a perfect square trinomial.
\({x^2} + \frac{3}{5}x + \frac{9}{{100}} = \frac{1}{5} + \frac{9}{{100}}\)	Add 3 to both sides of the equation to solve for \({\left( {x + 2} \right)^2}\)
\({\left( {x + \frac{3}{{10}}} \right)^2}\)=\(\frac{{29}}{{100}}\)	The perfect square trinomial is expressed as a cubed binomial.
\(\sqrt {{{\left( {x + \frac{3}{{10}}} \right)}^2}} = \sqrt {\frac{{29}}{{100}}} \ )	Take the square root of each side of the equation
\(x = – \frac{3}{{10}} \pm \frac{{\sqrt {29} }}{{10}}\)	Solve for \(x\).

The solutions of \(5{x^2} + 3x – 1 = 0\) are:
\(x = – \frac{{3 + \sqrt {29} }}{{10}},\; – \frac{{3 – \sqrt {29} }}{{10}}\)

The procedure used in the above equation will be used to find what is called the general formula for quadratic solutions.

General Formula of the Second Degree Equation.

General formula of quadratic equations

In this section we will find how to solve, in a general way, a quadratic equation

With \(a \ne 0\) let us consider the equation \(a{x^2} + bx + c = 0\).

\(a{x^2} + bx + c = a\left( {{x^2} + \frac{b}{a}x + \frac{c}{a}} \right) = 0\)

Since \(a \ne 0\) it is enough to solve:

\({x^2} + \frac{b}{a}x + \frac{c}{a} = 0\)

\({x^2} + \frac{b}{a}x + \frac{c}{a} = 0\)	Initial situation
\({x^2} + \frac{b}{a}x + \frac{c}{a} – \frac{c}{a} = – \frac{c}{a}\)	Add \( – \frac{c}{a}\) to each side of the equation.
\({x^2} + \frac{b}{a}x = – \frac{c}{a}\)	Since \(\frac{1}{2}\left( {\frac{b}{a}} \right) = \frac{b}{{2a}}\), by adding \({\left( { \frac{b}{{2a}}} \right)^2} = \frac{{{b^2}}}{{4{a^2}}}\) yields a perfect square trinomial.
\({x^2} + \frac{b}{a}x + \frac{{{b^2}}}{{4{a^2}}} = \frac{{{b^2}} }{{4{a^2}}} – \frac{c}{a}\)	The left hand side of the equation is a perfect square trinomial.
\({\left( {x + \frac{b}{{2a}}} \right)^2} = \frac{{{b^2} – 4{a^2}c}}{{4{ a^2}}}\)	The perfect square trinomial is expressed as a squared binomial. The algebraic fraction is done.
\(\sqrt {{{\left( {x + \frac{b}{{2a}}} \right)}^2}} = \sqrt {\frac{{{b^2} – 4{a^ 2}c}}{{4{a^2}}}} \)	Take the square root of each side of the equation.
\(\left| {x + \frac{b}{{2a}}} \right| = \frac{{\sqrt {{b^2} – 4{a^2}c} }}{{2a} }\)	Radical properties apply.
\(x + \frac{b}{{2a}} = \pm \frac{{\sqrt {{b^2} – 4{a^2}c} }}{{2a}}\)	Absolute value properties apply.
\(x + \frac{b}{{2a}} – \frac{b}{{2a}} = \pm \frac{{\sqrt {{b^2} – 4{a^2}c} } }{{2a}} – \frac{b}{{2a}}\)	To each side of the equation add \( – \frac{b}{{2a}}\) to solve for \(x\)
\(x = \frac{{ – b \pm \sqrt {{b^2} – 4ac} }}{{2a}}\)	The algebraic fraction is done.

The term \({b^2} – 4{a^2}c\) is called the discriminant of the quadratic equation \(a{x^2} + bx + c = 0\).

When the discriminant of the above equation is negative, the solutions are complex numbers and there are no real solutions. Complex solutions will not be covered in this note.

Given the quadratic equation \(a{x^2} + bx + c = 0\), if \({b^2} – 4{a^2}c \ge 0\). Then the solutions of this equation are:

\(\alpha = \frac{{ – b + \sqrt {{b^2} – 4ac} }}{{2a}}\)

\(\beta = \frac{{ – b – \sqrt {{b^2} – 4ac} }}{{2a}}\)

The expression:

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

It is called the General Formula of the quadratic equation.

Practical example 8: solve the equation \(3{x^2} – 2x – 5 = 0\)

\(to\)	\(b\)	\(c\)	Discriminating	real solutions
\(3\)	\( – 2\)	\( – 5\)	\({2^2} – 4\left( 3 \right)\left( { – 5} \right) = 4 + 60 = 64\)	\(x = \frac{{ – \left( { – 2} \right) \pm \sqrt {64} }}{{2\left( 3 \right)}} = \frac{{2 \pm 8} }{6}\)

The solutions of the equation are:
\(\alpha = – 1,\;\beta = \frac{5}{3}\)

Practical example 9: Solve the equation \( – 4{x^2} + 3x + 9 = 0\)

\(to\)	\(b\)	\(c\)	Discriminating	real solutions
\( – 4\)	3	9	\({3^2} – 4\left( { – 4} \right)\left( 9 \right) = 9 + 144 = 153\) \(153 = 9\left( {17} \right)\)	\(x = \frac{{ – 3 \pm \sqrt {9\left( {17} \right)} }}{{2\left( { – 4} \right)}} = \frac{{ – 3 \pm 3\sqrt {17} }}{{ – 8}}\)

The solutions of the equation are:
\(\alpha = \frac{{3 – 3\sqrt {17} }}{8},\;\beta = \frac{{3 + 3\sqrt {17} }}{8}\)

Practical example 10: Solve the equation \(5{x^2} – 4x + 1 = 0\)

\(to\)	\(b\)	\(c\)	Discriminating	real solutions
\(5\)	-4	\(1\)	\({\left( { – 4} \right)^2} – 4\left( 5 \right)\left( 1 \right) = 16 – 20 = – 4\)	Does not have

Miscellaneous Equations

There are non-quadratic equations that can be converted to a quadratic equation. We will see two cases.

Practical example 11: Finding the real solutions of the equation \(6x = 5 – 13\sqrt x \)

Making the change of variable \(y = \sqrt x \), the previous equation remains as:

\(6{y^2} = 5 – 13y\)

\(6{y^2} + 13y – 5 = 0\)

\(6{y^2} + 15y – 2y – 5 = 0\)

\(3y\left( {2y + 5} \right) – \left( {2y + 5} \right) = 0\)

\(\left( {2y + 5} \right)\left( {3y – 1} \right) = 0\)

Therefore \(y = – \frac{2}{5},\;\frac{1}{3}\).

Since \(\sqrt x \) only denotes positive values, we will only consider:

\(\sqrt x = \;\frac{1}{3}\)

Answer:

The only real solution is:
\(x = \frac{1}{9}\)

Worked example 12: Solve the equation \(\sqrt {\frac{x}{{x – 5}}} – \sqrt {\frac{{x – 5}}{x}} = \frac{5}{6 }\)

Making the change of variable:

\(y = \sqrt {\frac{x}{{x – 5}}} \)

We get the equation:

\(y – \frac{1}{y} = \frac{5}{6}\)

\(6{y^2} – 6 = 5y\)

\(6{y^2} – 5y – 6 = 0\)

\(6{y^2} – 9y + 4y – 6 = 0\)

\(3y\left( {2y – 3} \right) + 2\left( {2y – 3} \right) = 0\)

\(\left( {2y – 3} \right)\left( {3y + 2} \right) = 0\)

The possible values ​​of \(y\) are:

\(y = – \frac{2}{3},\;\frac{3}{2}\)

Of the above we will only consider the positive solution.

\(\sqrt {\frac{x}{{x – 5}}} = \frac{3}{2}\)

\(\frac{x}{{x – 5}} = \frac{9}{4}\)

\(4x = 9x – 45\)

\(5x = 45\)

\(x = 9.\)

The solutions are \(x = 9.\)