Question

Define quadratic polynomial.write one example.

Answer 1

In a quadratic equation, where you will have three terms : x² , x, coefficient. We will split the middle term to two terms, so that we can factorize the quadratic equation and find root.

So, Let's first know what is a Quadratic equation?

A quadratic equation is a polynomial of degree 2 , So the term with greatest degree would have a degree of 2 .

So , Basically, Quadratic equation can have 2 roots and in cases there might be 1 too.

Now, What's the general form of a quadratic equation?

The general form of a quadratic equation is ax² + bx + c.

So, Let us consider that β , α are the roots.

Thus x = α , β then ( x - α ) ( x - β ) is the quadratic polynomial.

Simplifying it as following ,

x ( x - β ) - α ( x - β )

x² - (α+ β)x + αβ

So , Comparing this with general equation.

ax² + bx + c = x² - (α + β)x + αβ

x² + b/a x + c/a = x² - (α+β)x + αβ

Comparing,

b/a = - (α + β)

c/a = αβ

So , If we are given a quadratic equation of dx² + ex + f, Middle term is mostly used when d = 1 , that is when you have to find

.roots having a product of f and having a sum of negative e.

As you have asked for a example let's see an example.

x² - 33x + 200

Here ,The numbers might seem large and basically you might be taking it hard to solve. but Basically it's just easy.

So, Here coefficient of x = 1 .

We should look for numbers whose product equals 200 .

So the pairs that do come into mind are 1 * 200 , 2 * 100 , 4 * 50 , 5 * 40 , 10 * 20 , 25 * 8 .

So, The next condition is the sum of numbers we choose must be negative the coefficient of x.

Here, Coefficient of x = -33

Negative of coefficient of x = 33 .

So , From the above pairs ( 25 , 8 ) have a sum of 33 .

Now, The big quadratic equation which seemed too tough looks like this

x² - ( 25 + 8 ) x + (25*8)

Which is looking very easily to solve, Let's factorise :)

x² -25x - 8x + 300

x ( x - 25 ) -8 ( x - 25)

( x - 25 ) ( x - 8 )

So, the above are the factors.

if you want zeroes, Just equate them with zero

x - 25 = 0 , x = 25

x - 8 = 0 , x = 0 .

So, We must look for the factor pairs of the coefficient of constant, so that problem is easy.

There are quadratic equations like 3x² + 6x - 9 where Coefficient of x² isn't 1 , in that case, divide the entire polynomial by coefficient of x² and after factorising multiply one of the factor by the coefficient of x²

The above polynomial becomes x² + 2x - 3

x² + 2x - 3

= x² + 3x - x - 3

= x ( x + 3 ) -1 ( x - 3 )

= ( x - 1 )( x - 3)

Multiply any of factor by 3 .

= (3x - 3 ) ( x - 3 )

= ( x - 1 ) ( 3x - 9 )

Which ever the factor, you multiply won't change the roots. So no worries in that.

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Quadratic polynomial : A polynomial whose degree is 2 is known as quadratic polynomial or An algebraic expression whose degree is 2 and the powers of variables involved in the polynomial are non negative integral powers.

Example : 2x² , 3x² + 5x + 3 , 6x² - 7 are quadratic polynomials . x² + 1/x is not a polynomial.

Answer 2

Solution :-

Quadratic Equation -

A quadratic equation is an equation of the second degree. It means that it contains at least one term that is squared. The standard form is ax² + bx + c = 0 with a, b and c being constants or numerical coefficients, and x is an unknown variable. One absolute rule is that the first constant "a" cannot be a zero.

Here are some examples of quadratic equations in the standard form (ax² + bx + c = 0)

1) 6x² + 11x - 35 = 0

2) 2x² - 4x - 2 = 0

3) - 4x² - 7x + 12 = 0

4) 20x² - 15x - 10 = 0

5) x² - x - 3 = 0

6) 5x² - 2x - 9 = 0

7) 3x² + 4x + 2 = 0

8) - x² + 6x + 18 = 0