Question

Find the zeroes of the quadratic polynomial 5x2 + 8x – 4 and verify the relationship between the zeroes and the coefficients of the polynomial.

Answer 1

Answer :-

★ Concept :-

Here the concept of zeroes of a polynomial has been used. According to this, the zeroes of polynomial α and β are given, as

$\longrightarrow \: \: \: \: {\bf{\blue{\alpha \: + \: \beta \: = \: \dfrac{(-b)}{a}}}}$

And,

$\longrightarrow \: \: \: \: {\bf{\blue{\alpha \: \times \: \beta \: = \: \dfrac{c}{a}}}}$

where, a is the coefficient of first term, b is the coefficient of second term and c is the third constant term in a polynomial of form ax² + bx + c

★ Question :-

Find the zeroes of the quadratic polynomial 5x² + 8x – 4 and verify the relationship between the zeroes and the coefficients of the polynomial.

★ Solution :-

$\:\:\:{\boxed{\rm{\red{Finding \: The \: Zeroes \:}}}}$

Given,

➫ p(x) = 5x² + 8x - 4

This equation can be easily solved using Splitting the middle term method. So,

➥ 5x² + 8x - 4 = 0

(Since, p(x) = 0)

➥ 5x² + 10x - 2x - 4 = 0

Taking like terms in common, we get,

➥ 5x(x + 2) - 2(x + 2) = 0

➥ (5x - 2)(x + 2) = 0

Here, either (5x - 2) = 0 or (x + 2) = 0. So,

➥ 5x - 2 = 0 or x + 2 = 0

➥ 5x = 2 or x = -2

Hence,

$\bold{x_{{\pink{(zeroes \: of \: the \: polynomial)}}}} \: = \dfrac{2}{5} \: \: or \: \: -2$

______________________________

$\:\:\:{\boxed{\sf{\green{Verification \: of \: the \: relationship \: between \: the \: zeroes}}}}$

Here,

$\alpha \: = \: \dfrac{2}{5} \\\\And, \beta \: = \: -2$

Also, a = 5 , b = 8 and c = -4

According to the given concept, we get,

~ Case I :-

$\Longrightarrow\: \: \: \: {\alpha \: \: + \: \: \beta \: = \: \dfrac{(-b)}{a}}$

$\Longrightarrow \: \: \: \: {\dfrac{2}{5} \: + \: (-2) \: = \: \dfrac{(-8)}{5}}$

$\Longrightarrow\: \: \: \: {\dfrac{(2 - 10)}{5} \: = \: \dfrac{(-8)}{5}}$

$\Longrightarrow \: \: \: \: \dfrac{(-8)}{5} \: = \: \dfrac{(-8)}{5}$

Clearly, LHS = RHS

~ Case II :-

$\Longrightarrow \: \: \: \: {\bold{\alpha \: \times \: \beta \: = \: \dfrac{c}{a}}}$

$\Longrightarrow \: \: \: \: {\dfrac{2}{5} \: \times \: (-2) \: = \: \dfrac{(-4)}{5}}$

$\Longrightarrow\: \: \: \: {\dfrac{(-4)}{5} \: = \: \dfrac{(-4)}{5}}$

Clearly, LHS = RHS

Here both the conditions satisfy. So our answer is correct.

Hence verified.

_____________________

★ More to know :-

• Zeroes of the polynomial can be found out by using :-

1. Splitting the Middle Term Method
2. Perfect Square Method
3. Quadratic Formula

• Polynomial is equation formed by using both constant and variable term.

• Polynomials can be classified as :-

1. Linear Polynomial
2. Quadratic Polynomial
3. Cubic Polynomial
4. Bi - Quadratic Polynomial

• If we take example of any polynomial like ax² + bx + c = 0 , then here a, b and c is the constant term and x is the variable term.

Answer 2

Solution:

Given quadratic polynomial is

• 5 x^2 + 8 x - 4

Let, us first find its zeroes

So, equating the given polynomial with zero

→ 5 x^2 + 8 x - 4 = 0

→ 5 x^2 + ( 10 - 2 ) x - 4 = 0

→ 5 x^2 + 10 x - 2 x - 4 = 0

→ 5 x ( x + 2 ) - 2 ( x + 2 ) = 0

→ ( 5 x - 2 ) ( x + 2 ) = 0

so, 5 x - 2 = 0

implies, x = 2/5  

and, x + 2 = 0

implies, x = -2

Therefore, two zeroes of given quadratic polynomial are:

• 2/5 and -2.

Now,

As we know for a given quadratic polynomial of the form ax^2 + bx + c

with two zeroes

• Sum of zeroes = - (coefficient of x ) / (coefficient of x^2)
• and, product of zeroes = ( constant term ) / ( coefficient of x^2 )

So,

In the given quadratic polynomial 5 x^2 + 8 x - 4

• coefficient of x^2 = 5
• coefficient of x = 8
• constant term = -4

So,

→ Sum of zeroes = - ( 8 ) / 5

→ (2/5) + (-2) = -8/5

→ ( 2 - 10 ) / 5 = -8/5

→ -8/5 = -8/5

LHS = RHS

hence, verified.

also,

→ Product of zeroes = ( -4) / ( 5 )

→ ( 2/5 ) · ( -2 ) = -4/5

→ -4/5 = -4/5

LHS = RHS

hence, verified.