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Solving Equations

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Solving equations is the process of finding the values of the variables or unknown quantities that are satisfied by the condition defined in the equation. An equation is described as the combination of the variables, constants and operations. These equations are solved by using different methods. Solving an equation is just like solving a puzzle. There are several equations that define the different types of equations.

Examples:
3s - 8 =  5s - 6 + p
$\frac{6}{7}$a + $\frac{5}{6}$ = 5a - $\frac{125}{9}$.


The above equations have variables like a, s, p and the constant values are joined together using some mathematical operations. While solving an equation, there are several methods you can adopt, like combining like terms, adding or subtracting same value on both sides, factoring, expansion, dividing terms by non zero value, difference of squares etc.

Solving Quadratic Equations

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The standardized form of a quadratic equation is ax$^{2}$ + bx + c = 0, where a, b and c can have any value and "a" cannot be zero. When the equation is in a standard form, larger values of "a" will shrink the curve, while smaller values of "a" expand it and negative values flip the curve upside down.

Quadratic equation is a univariate polynomial equation of the second degree. The constants a, b, and c are called quadratic coefficients or constant terms.

They can be solved by factoring, completing the square, using the quadratic formula, and graphing. Given below is an example of a quadratic equation.

Let us solve 5x$^{2}$ + 3x + 2 = 0 by using quadratic formula.

In the given equation, coefficients of a, b and c are 5, 3 and 2 respectively.
Formula for quadratic equation is as follows:

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

Plug in the values for a, b,and c into the quadratic formula

x = $\frac{-3\pm\sqrt{3^{2}-4 \times 5 \times 2}}{2 \times 5}$

x = $\frac{-3\pm\sqrt{-31}}{10}$

We see that we can have two values for x, let they be $x_{1}$ and $x_{2}$

$x_{1}$ = $\frac{-3+\sqrt{-31}}{10}$

= $\frac{-3}{10}+i\frac{1}{10}\sqrt{31}$

$x_{2}$= $\frac{-3-\sqrt{-31}}{10}$

= $\frac{-3}{10}-i\frac{1}{10}\sqrt{31}$

Therefore, the values of $x_{1}$ and $x_{2}$ are $\frac{-3}{10}+i\frac{1}{10}\sqrt{31}$ and
$\frac{-3}{10}-i\frac{1}{10}\sqrt{31}$ respectively.

Solving Linear Equations

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Linear equation is a mathematical expression that has an equal sign and linear expressions. They are the simplest equations that you'll deal with. A linear equation cannot have exponents, multiplying and dividing a variable, a variable under a square root. A linear function is always represented by a straight line and plotting linear functions is very easy.

Any function of the form f(x) = mx + b, m $\neq$ 0 is called a linear function.
Let us solve for x in x - 7 = 5.

Given: x - 7 = 5
Adding 7 to both sides of the equation, we get
x - 7 + 7 = 5 + 7
x = 12
The answer is x = 12.

Solving Logarithmic Equations

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Logarithm of a number x with respect to base b is the exponent by which b must be raised to yield x. That is, a$^{x}$ = M. Given below is an example for solving logarithmic equations.

Let us solve for x in the equation 5 log(3x) = 19

Divide both sides of the given equation by 5.

log(3x) = $\frac{19}{5}$

Now, convert the logarithmic equation to an exponential equation.

The above equation can be written as

10$^{\frac{19}{5}}$ = 3x

Dividing both sides of the above equation by 3, we get

x = $\frac{10^{\frac{19}{5}}}{3}$

x = 2103.1911

Solving Rational Equations

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A rational equation will be in the form y = $\frac{N(x)}{D(x)}$ where, N and D are polynomials. It looks like a fraction. Rational equations are equations containing rational expressions.

Let us solve $\frac{x - 2}{3}$ + $\frac{x}{2}$ = 4

Given $\frac{x - 2}{3}$ + $\frac{x}{2}$ = 4

The least common denominator of the fractions is 6.

$\frac{2(x-2)+3x}{6} = 4$


By cross multiplying, we get
2(x - 2) + 3x = 24
2x - 4 + 3x = 24
5x = 28

x = $\frac{28}{5}$

x = 5.6

Solving Exponential Equations

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In an exponential equation, variable occur in the exponent.

If the bases are the same, set the exponents equal.

Let us solve $5^{(2x - 3)}$ = 125
Given $5^{(2x - 3)}$ = 125
$5^{(2x - 3)} = 5^{3}$
Equating exponents, as bases are same.
2x - 3 = 3
2x = 6
$\Rightarrow$ x = 3

Solving Absolute Value Equations

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An absolute value function is a function that contains an algebraic expression within absolute value symbols. Recall that the absolute value of a number is its distance from 0 on the number line. It is defined as f(x) = |x|.

|x| is pronounced as "mod of x" or "absolute value of x".

Let us find the possible solutions for the equation y = |x + 5|.

Plugging different values of x, we get the values of y as shown in the below table:

x y = |x + 5|
-10 |-10 + 5| = |-5| = 5
-9 |-9 + 5| = |-4| = 4
-6 |-6 + 5| = |-1| = 1
0 |0 + 5| = |5| = 5
2
|2 + 5| = |7| = 7

Solving Radical Equations

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Radical functions contain functions involving roots. A radical function contains a radical expression with the independent variable in the radicand. An example of radical function is y = $\sqrt{x}$.

Domain of the function should not have a negative value under the square root sign. Otherwise, it will lead to complex number.

Let us solve 5 = $\sqrt{3x}$ + 12
Isolate the radical by subtracting 12 from both the sides.
5 - 12 = $\sqrt{3x}$ + 12 - 12

-7 = $\sqrt{3x}$

Squaring on both sides, we get
49 = 3x
Simplifying, we get x = 16.33

Solving Polynomial Equations

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A polynomial function is a function which can be written in the form f(x)= $a_{n}x^{n} + a_{n-1}x^{n-1}+......+ a_{1}x + a_{0}$, where $a_{n} \neq 0$ and $a_{n},a_{n-1}.......a_{0}$ are real numbers and n being a non-negative integer. A polynomial can be expressed in terms that only have positive integer, exponents and the operations of addition, subtraction, and multiplication. The above can further be categorized on the basis of number of terms and the degree of the polynomial.

The degree of a polynomial is the highest power which is raised to the variable.

Let us solve 4x$^{2}$ + 3x + 5 = 0 by using completing the square method.

Step 1:
Divide the given equation by 4.

x$^{2}$ + $\frac{3}{4}$ x + $\frac{5}{4}$ = 0

Step 2:

Move the constant to the right keeping all x terms containing x on one side.

x$^{2}$ + $\frac{3}{4}$ x = - $\frac{5}{4}$

Step 3:

Find half of the x-term coefficient and square it. Add the obtained value in both the sides.

x-term coefficient = $\frac{3}{4}$

Half of the x term coefficient = $\frac{3}{8}$

After Squaring, we get ($\frac{3}{8})^{2}$ = $\frac{9}{64}$

Add $\frac{9}{64}$ to both the sides,

x$^{2}$ + $\frac{3}{4}$x + $\frac{9}{64}$ = - $\frac{5}{4}$ + $\frac{9}{64}$

Step 4 :

Simplifying the right side, we get

x$^{2}$ + $\frac{3}{4}$x + $\frac{9}{64}$ = -$\frac{71}{64}$

Step 5:

Write the perfect square on the left.
(x + $\frac{3}{8}$)$^{2}$ = -$\frac{71}{64}$

Step 6 :

Take the square root of both the sides.

x + $\frac{3}{8}$ = $\pm$$\sqrt{\frac{71}{64}}$

Step 7:


x = - $\frac{3}{8}$ $\pm$ $\sqrt{-\frac{71}{64}}$

Therefore, $x_{1}$ = -$\frac{3}{8}$ + $i\frac{1}{8}$ $\sqrt{71}$ and $x_{2}$ = -$\frac{3}{8}$ - $i\frac{1}{8}$ $\sqrt{71}$

Solving Trigonometric Equations

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Equations involving trigonometric functions of the unknowns can be solved by using simple algebraic ideas and trigonometric identities. sin$^{3}$x - 5x is difficult to solve. Given below is an example of a trigonometric equation.

Given Cos $\theta$ = $\frac{15}{19}$, let us find Sec $\theta$.

As sine and cosine are reciprocal functions, we have

Sec $\theta$ = $\frac{1}{\cos \theta}$

= $\frac{19}{15}$