Question

if a and b are the zeroes of polynomial x ^2 + x - 2 find a polynomial whose zeroes are 3a+ 1 and 3b + 1 ​

Answer 1

Answer:

x² + x - 20

Step-by-step explanation:

Given,

a , b are the zeroes of the polynomial of 'x² + x - 2'

To Find :-

The quadratic polynomial whose zeroes are '3a + 1' and '3b + 1'

How To Do :-

Here they given the quadratic polynomial and said to us that a , b are zeroes of that and we were asked to find the quadratic polynomial whose zeroes are '3a + 1' and '3b + 1'. So first of all we need to find the zeroes of the quadratic polynomial by using the quadratic formula. After finding that we need to equate them 'a , b' and we need to find those values. After obtaining that we need to find the values of '3a + 1 and 3b+1' and we need to substitute them in the quadratic polynomial formula.

Formula Required :-

Quadratic formula :-

$x=\dfrac{-B\pm\sqrt{B^2-4AC}}{2A}$

Quadratic polynomial using zeroes :-

x² - (sum of zeroes)x + (product of zeroes).

Solution :-

x^2 + x - 2

→ A = 1

B = 1

C = -2

Substituting in the quadratic formula :-

$x=\dfrac{(-1)\pm\sqrt{(1)^2-4(1)(-2)}}{2(1)}$

$=\dfrac{-1\pm\sqrt{1+8}}{2}$

$=\dfrac{-1\pm\sqrt{9}}{2}$

= -1 ± 3/2

= -1+3/2 , -1-3/2

=  2/2 , -4/2

= 1 , -2

∴ zeroes of the quadratic polynomial = (a , b)

(-2 , 1) = (a , b)

→ a = -2

b = 1

3a + 1 = 3(-2) + 1

= -6 + 1

= -5

3b + 1 = 3(1) + 1

= 3 + 1

= 4

∴ zeroes of the new quadratic polynomial = -5 , 4

Sum of the zeroes = -5 + 4

= -1

Product of the zeroes = -5 × 4

= -20

Substituting in the formula :-

= x² - (-1)x - 20

= x² + x - 20

∴ The required quadratic polynomial = x² + x - 20.

Answer 2

Step-by-step explanation:

Answer

As we know the sum of the roots of a quadratic equation is equal to the negation of the coefficient of the second term, divided by the leading coefficient. The product of the roots of a quadratic equation is equal to the constant term (the third term), divided by the leading coefficient.

Since a and b are the zeros of the quadratic polynomial f(x)=2x

2

−5x+7

∴a+b==

2

−(5)

=

2

5

and ab=

2

7

Let s and p denote respectively the sun and product of the zeros of the required polynomial,

Then S = (2a+3b)+(3a+2b)=5(a+b)=5×

2

5

=

2

25

And p=(2a+3b)(3a+2b)

⇒p=6a

2

+6b

2

+13ab=6a

2

+6b

2

+12ab+ab

=6(a

2

+b

2

+2ab)+ab=6(a+b)

2

+ab

⇒p=6×(

2

5

)

2

+

2

7

=

2

75

+

2

7

=41

Hence, the required polynomial g(x) is given by

g(x)=k(x

2

−Sx+p)

or g(x)=k(x

2

−

2

25

x+41) where k is any non-zero real number