An expression which consists of constant, variables and exponent values is known as polynomial expression. The polynomial expression is combined with the addition, subtraction, and multiplication. Like $5x^{2} + 9x + 12$ = $0$; This given expression is a polynomial expression and one exponent value is present in this expression. The expression which contains a Square root values is known as radical expression. For example: $\sqrt{6},\ 3 \sqrt{x + 2},\ 8 \sqrt{16}$ and $6 + 3 \sqrt{12}$; these all are radical expressions. The symbol $\sqrt{}$ is used to represents the square root or $n$ th roots of the expression. The other definition, an expression which contains the radical expression with the variable in the radicands is also known as radical equation. Here we will be exploring Polynomials and radical expression. Suppose we have a radical expression $\sqrt{\frac{x^{4}17}{y5}}$ Where the value of ‘$x$’ is $3$ and the value of ‘$y$’ is $9$ and we have to solve the radical expression; For solving this expression we follow some steps: The given expression is $\sqrt{\frac{x^{4}17}{y5}}$ And value of ‘$x$’ and ‘$y$’ is $3$ and $9$; Step 1: First substitute the values of ‘$x$’ and ‘$y$’ in the given radical expression. $\Rightarrow$ $\sqrt{\frac{(3)^{4}  17}{95}}$ Step 2: Then we find the positive square root. $\Rightarrow$ $\sqrt{\frac{64}{4}}$ Step 3: Then solve the expression. $\Rightarrow$ $\sqrt{16}$ $\Rightarrow$ $4$; Where $x$ = $3$ and $y$ = $9$ Now we will see how to solve polynomial expression; Suppose we have $3x^{2} 5x + 3$ at $x  4$; For solving this expression we have to follow some steps: Step 1: First we have a polynomial expression. Step 2: Then put the value of ‘$x$’ as $4$. $\Rightarrow$ $3x^{2} 5x + 3$; On putting the value of $x$ = $4$ we get: $\Rightarrow$ $3(4)^{2} 5(4) + 3$; Step 3: Then we solve the expression: $\Rightarrow$ $3 \times 16 + 20 + 3$; $\Rightarrow$ $48 + 23$; $\Rightarrow$ $71$;

Quadratic equation can be defined as polynomial equation with highest degree $2$. Quadratic equation can also be called as a second degree equation. Consider the following function $f (X)$ = $J X^{2} + KX + L$, Since highest degree of variable '$X$' is two $(2)$ hence it is called as Quadratic Equation. $J, K$ and $L$ are quadratic coefficients. Lets see how to Graph Quadratic Equation.
Domain of the above function $f (X)$ is the Set of all Real Numbers. $y$intercept of graph of $f (X)$ is given by $f (0)$ = $c$ and $x$intercepts are obtained by solving the equation $JX^{2} + KX + L$ = $0$ using the Quadratic Formula.
Lets understand the process of analyzing graphs of quadratic Functions. Range of quadratic function $f (X)$ can be found as shown below:
First, $f (X)$ = $J (X  h)^{2} + C$,
$f (X)$ = $JX^{2}  2JhX + Jh^{2} + C$
Two conditions must be followed $2Jh$ = $K$ (first equation) and $Jh^{2} + C$ = $L$ (second equation). From first equation, $h$ = $\frac{K}{2J}$. Using second equation,
$C$ = $\frac{L  K^{2}}{4J}$
The term $(X  h)^{2}$ is either positive or zero, hence $(X  h)^{2} \geq 0$.
If $J > 0$, multiply both sides of inequality above by $J$
$J(X  h)^{2} \geq 0$,
Add $C$ to both sides of inequality
$J(X  h)^{2} + C \geq C$,
Thus $f (X) \geq C$
Thus range of $f (X)$ is given by
$(C, + \infty)$. Since minimum value of $f (X)$ is $C$ and it can be concluded that range of $f (X)$ can also be given by $( \infty, C)$.
From above discussion, it is clear that graph of quadratic function is called as Parabola and Point with coordinates $(h, C)$ is called vertex of parabola.
Consider the equation $f (x)$ = $2x^{2} + 2 x  4$, for which graph looks like this:
Quadratic equation is an equation whose highest degree is two. Degree $2$ indicates that highest power in equation is two.
Let’s discuss process of solving Quadratic Equation by factoring. Let’s consider the following quadratic equation to understand how to solve quadratic equations by factoring?
$A\ y^{2} + B\ y + C$ = $0$,
Since maximum degree of variable '$y$' is two $(2)$, hence this type of equation is referred to quadratic equation. $A, B$ and $C$ are called quadratic coefficients. There are two methods to solve a quadratic equation. One is direct Factorization and second is using Quadratic Formula.
Above equation can be solved using quadratic formula which is given as:
$y_{0}$ = $\frac{B \pm \sqrt{b^{2}  4AC}}{2A}$
This is the standard formula for factorizing quadratic equation.
Let’s take following example:
Suppose we are given $P^{2} + 9P + 20$ = $0$ then what is the solution of this equation? There are two types of factorization methods. One method includes direct factorization. In direct factorization, we multiply first $(P^{2})$ and last $(20)$ terms which gives a new value $(20P^{2})$. Then distribute the middle value $(9P)$ in such a way that its multiplication gives $(20P^{2})$ and addition or subtraction gives middle value. Another method is use of quadratic formula.
Using the first method, we get following solution:
$P^{2} + 9P + 20$ = $P^{2} + 5P + 4P + 20$ = $0$,
$P (P + 5) + 4 (P + 5)$ = $0$,
$(P + 5) (P + 4)$ = $0$,
Then solution will be,
$P$ = $5$ or $P$ = $4$,
Using quadratic formula same results can be obtained. This is all about factoring quadratic equation.
Standard Quadratic Equation is written in form $ax^{2} + b x + c$ = $0$, but in case of quadratic inequalities expression is written in as,
1) $ax^{2} + b\ x + c > 0$, example: $x^{2} + 4x > 5$,
2) $ax^{2} + b\ x + c < 0$, example: $8x^{2} < 29$,
3) $ax^{2} + b\ x + c < 0$, example: $6 \geq x^{2} – x$,
4) $ax^{2} + b\ x + c \leq 0$, example: $4y^{2} + 1$ = $8y$
There are some steps for solving quadratic inequalities :
Step 1: First of all we have to move all terms to one side.
Step 2: Factorize the inequality.
Step 3: After this find the roots of this quadratic equation.
Step 4: We can graph quadratic inequalities by plotting coordinates on graph and then shading appropriate area according to inequality sign.
Let’s take an example: $x^{2} + 4x < 5$,
Solution:
First of all we will move $5$ to left side
$x^{2} + 4x – 5 < 0$, now solve this equation and find the root, so, $x^{2} + 5x  x  5 < 0 \Rightarrow x (x + 5)  2 (x + 5) < 0$ ,
$\Rightarrow\ (x  2) (x + 5) < 0$, so $x$ = $2,\  5$.
Let’s take the value $x$ = $0$ then $0 – 0 5 < 0$
Now $x$ = $1, 1 + 4 5 < 0$
Now $x$ = $4 \Rightarrow 16 – 16 5 < 0$
Now $x$ = $3 \Rightarrow 9 – 12 – 5 < 0$
So it is proved that $x^{2} + 4x – 5 < 0$ in interval $(2 5)$.
Quadratic equation is written as $ax^{2} + bx + c$ = $0$, from this equation we can calculate Quadratic Formulas and Discriminant. Quadratic
formula is given as: $\frac{(b\ \mp\ \sqrt{(b^{2}\ –\ 4ac)})}{2a}$, this formula is called, this formula gives us two solutions,
$x_{1}$ = $\frac{(b\ +\ \sqrt{(b^{2}\ –\ 4ac)})}{2a}$ and $x_{2}$ = $\frac{(b\ \ \sqrt{(b^{2}\ –\ 4ac)})}{2a}$.
Similarly we can find the Discriminant, discriminant provides the information about nature of roots and solution of any Quadratic Equation. Discriminant formula is given by:
= $\frac{(4 + 2 i)}{2}$, $\frac{(4  2 i)}{2}$,
formula is given as: $\frac{(b\ \mp\ \sqrt{(b^{2}\ –\ 4ac)})}{2a}$, this formula is called, this formula gives us two solutions,
$x_{1}$ = $\frac{(b\ +\ \sqrt{(b^{2}\ –\ 4ac)})}{2a}$ and $x_{2}$ = $\frac{(b\ \ \sqrt{(b^{2}\ –\ 4ac)})}{2a}$.
Similarly we can find the Discriminant, discriminant provides the information about nature of roots and solution of any Quadratic Equation. Discriminant formula is given by:
Discriminant = $b^{2}\ –\ 4ac$.
Suppose a quadratic equation $x^{2}\ +\ 4x\ +\ 4$ = $0$, here $a$ = $1,\ b$ = $4,\ c$ = $4$ so discriminant is $4^{2}\ –\ 4ac\ \Rightarrow\ 16\ –\ 4\ .\ 1\ .\ 4\ \Rightarrow\ 16\ \ 16$ = $0$.
We can also find the nature of solution using this discriminant .
Case 1: If value of $b^{2} – 4 a c > 0$ means discriminant is positive discriminant, in this case two real solutions are generated, if discriminant is a perfect Square then roots are rational, otherwise they are irrational.
Case 2: If value of discriminant is zero, $b^{2} – 4 a c$ = $0$, then there is only one real solution.
Case 3: If value of discriminant is negative, $b^{2} – 4 a c < 0$, then in this case there is no real solution, there will be two imaginary solutions.
Example:
Calculate the discriminant, nature and number of solutions of equation $y$ = $x^{2} + 4x + 5$, in this equation $a$ = $1,\ b$ = $4,\ c$ = $5$, so discriminant = $b^{2} – 4 a c?$
Solution:
Solution:
Put value of $a, b, c$ in this formula, then $(42 – 4 .1.5) \Rightarrow\ 16 – 20$ = $4$.
Here discriminant is negative, so there are no real solution for this quadratic equation, solutions are imaginary. So using Quadratic Formula for
$y$ = $x^{2} + 4x + 5,$
$X_{1}$ = $(4) +$ $\frac{\sqrt{(4^{2} – 4 .1.5)})}{2.1}$
= $\frac{(4  \sqrt{(4)})}{2}$,
= $\frac{(4 + 2 i)}{2}$, $\frac{(4  2 i)}{2}$,
= $2 + i, 2  i$
An equation with highest degree of $2$ is called as Quadratic Equation. We can say that an equation in which highest power is a Square is known as quadratic equation. If highest power is more than $2$ then it is not a quadratic equation. It can be written as: $ax^{2} + bx + c$ = $0$ and Quadratic Formula is given by:
$\Rightarrow\ x$ = $ b + $\frac{\sqrt{(b^{2}  4ac)}}{2a}$, its alternate form is given by:
$\Rightarrow\ x$ = $\frac{2c}{ b}$ + $\sqrt{(b^{2}  4ac)}$.
Lets understand how to solve quadratic equation by Graphing. To solve quadratic equations by graphing method we need to follow some steps:
Step 1: Let quadratic equation is $p^{2}  8p + 15$ = $0$.
Step 2: Now with help of above defined formula we can find the factors. As we know that quadratic formula is given as:
$\Rightarrow\ x$ = $ b$ + $\frac{(b^{2}  4ac)}{2a}$. In equation value of '$a$' is $1$, value of '$b$' is $8$ and value of '$c$' is $15$.
$\Rightarrow\ x$ = $ b$ + $\frac{(b^{2}  4ac)}{2a}$. In equation value of '$a$' is $1$, value of '$b$' is $8$ and value of '$c$' is $15$.
Step 3: Put these values in formula. On putting these values we get:
$\Rightarrow\ x$ = $ b +$ $\frac{\sqrt{(b^{2}  4ac)}}{2a}$,
$\Rightarrow\ x$ = $ (8)$ + $\frac{\sqrt{((8)^{2}  4 \times 1 \times 15)}}{2 (1)}$,
$\Rightarrow\ x$ = $8$ + $\frac{\sqrt{(64  60)}}{2}$,
$\Rightarrow\ x$ = $8$ + $\frac{\sqrt{4}}{2}$, so here we will get two factors of given equation.
$X$ = $4 + \sqrt{1}$ and $X$ = $4  \sqrt{1}$. Now using these factors we can plot quadratic equation graph. The quadratic equation are shown below:
This is all about quadratic equation graph. This is the process of Solving Quadratic Equations by graphing.
Quadratic equation is a trinomial that means this equation has three terms. Standard form of Quadratic Equation is $ax^{2} + bx + c$. Here $a, b$ and $c$ are constants. Let us discuss the process of Solving Quadratic Equations by completing squares.
In mathematics completing the Square means to find the last term of perfect square trinomial. When we square the binomial term we get a trinomial expression. Meaning of squaring the binomial term is that we multiply the term to it. Let us see some examples of perfect trinomial square:
If we have a binomial as $x + 1$, when we find the square of binomial term we get a trinomial expression which is $(x + 1)^{2}$ = $x^{2} + 2x + 1$.
In order to complete the square we will find the square of binomial. There is a particular formula to get the square of binomial.
$(a + b)^{2}$ = $a^{2} + 2ab + b^{2}$,
General form of quadratic equation is $ax^{2} + bx + c$.
When we compare these two equations we find that '$a$' is a variable and '$b$' is a Integer value. '$C$' is half of square of '$b$'.
$(x + 3)^{2}$ = $x^{2} + 2(3)x + 3^{2}$,
$(x + 3)^{2}$ = $x^{2} + 6x + 9$
Here '$a$' is a variable and '$b$' is an integer. Value of $c$ = $\frac{1}{2}$ $(b)2$ which is $\frac{1}{2}$ $(3)2$ = $9$.
General formula of a perfect square trinomial is:
$(x + \frac{p}{2})^{2}$ = $x^{2} + 2$ $(\frac{p}{2})$ $x +$ $(\frac{p}{2})^{2}$.
This is the method by which we can complete a square.