Question

Find the quardratic polynomial whose zeroes are 2 and_-6.verify the relation between the coefficent and the zeroes of the polynomial

Answer 1

$\huge\mathfrak\green{Answer:}$

Given:

• We have been given two zeroes of a quadratic polynomial as 2 and -6.

To Find:

• We need to find the quadratic polynomial and verify the relationship between zeroes and coefficients of the polynomial.

Solution:

We have been given two zeroes of a quadratic polynomial as 2 and -6. Therefore, α = 2 and β = -6.

Now, Sum of zeroes (α + β) =

2 + (-6)

= 2 - 6

= -4__________(1)

Product of zeroes (αβ) =

2 × (-6)

= -12_________(2)

Now, we can find the polynomial by this formula:

k[x² - (α + β)x + (αβ)]

Substituting the values from equation 1 and 2, we have

k[x² - (-4)x + (-12)]

=> k[x² + 4x - 12]

= x² + 4x - 12

Hence the required polynomial is x² + 4x - 12.

Now, we need to verify the relationship between zeroes and coefficients of this polynomial.

Sum of zeroes = -b/a

= -4/1

= -4

Product of zeroes = c/a

= -12/1

= -12

Hence the relationship between the zeroes and coefficients of this polynomial is verified.

Answer 2

Provided with zeroes of the quadratic polynomial,we are supposed to find the quadratic polynomial initially whose zeroes would be 2 and -6. And henceforth verifying the relation between the coefficient and zeroes of the polynomial.

Let the two zeroes of the polynomial be $\sf{\alpha and \beta}$

Let the value of $\alpha$ be 2 and the value of $\beta$ be -6.

Now, find the sum of the zeroes.

=> $\sf{Sum\:of\: zeroes\:=\:\alpha\:+\:\beta}$

=> $\sf{Sum\:of\:zeroes\:=2\:+(-6)}$

=> $\sf{Sum\:of\: zeroes\:=2-6}$

=> $\sf{Sum\:of\:zeroes\:=-4....(1)}$

Now, move on to product of zeroes.

=> $\sf{Product\:of\: zeroes\:=\:\alpha\:\times\:\beta}$

=> $\sf{Product\:of \: zeroes\:=2\:\times\:-6}$

=> $\sf{Product\:of\:zeroes\:=-12....(2)}$

Now we know that we can use values of equations (1) and (2) to form our quadratic polynomial.

We know we can form the polynomial using,

• $\sf{x^2\:-\:(sum\:of\:zeroes)x\:+\:(Product\:of\: zeroes)}$

=> $\sf{x^2\:-(-4)x+(-12)}$

=> $\sf{x^2-4x-12}$

Now we have the desired quadratic polynomial with zeroes as 2 and -6.

Now verify the relation between the coefficient and zeroes.

=> $\sf{Sum\:of\:roots\:=\:\alpha\:+\:\beta}$

=> $\sf{Sum\:of\:roots\:=\:2+(-6)}$

=> $\sf{Sum\:of\:roots\:=\:2-6}$

=> $\sf{Sum\:of\:roots\:=-4....(3)}$

From equation (1) and (3),we satisfy the relation between the sum of roots and the coefficient x.

Now, product of roots.

$\sf{Product\:of\:zeroes\:=\:\alpha\:\times\:\beta}$

$\sf{Product\:of\:zeroes\:=\:2\:\times\:(-6)}$

$\sf{Product\:of\: zeroes\:=-12....(4)}$

From (2) and (4) we satisfy the relation between the product of zeroes with constant term.

Peace! :)