Question

solve the 19 th one plss​

Answer 1

Question:

If the zeros of the polynomial f(x)=x²+5x-4 are @ and ß,find the polynomial with the zeros (@+2)/3 and (ß+2)/3.

Solution:

Given that (@+2)/3 and (ß+2)/3 are the zeros of the required polynomial

We have,

f(x)=x²+5x-4

Here,

★Sum of zeros: -x coefficient/x²coefficient

@+ß= -5

★Product of zeros:

constant term/x²coefficent

@ß= -4

Let S and P represent the sum and product of zeros of required polynomial

respectively

★The zeros would be:

(@+2)/3= (-5+2)/3= -3/3

(ß+2)/3= (-4+2)/3= -2/3

Now,

S= (@+2)/3+(ß+2)/3

= -3/3-2/3

= -5/3

Also,

P= [(@+2)/3][(ß+2)/3]

=(-3/3)(-2/3)

= 6/9

Required Polynomial:

x²-Sx+P

=x²-(-5/3)x+6/9

Multiplying the whole equation by 9

= 9x²+15x+6

Answer 2

given:- alpha and beta are the zeroes of polynomial x² + 5x - 4

let's find the zeroes by splitting the middle term method.

➡ x² + 5x - 4 = 0

➡ x² + (4x - x) - 4 = 0

➡ x² + 4x - x - 4 = 0

➡ x(x + 4) - 1(x + 4) = 0

➡ (x + 4) (x - 1) = 0

• x = -4

or

• x = 1

hence, alpha = -4 and beta = 1

now, we have to find a quadratic equation whose zeroes are (α + 2)/3 and (β + 2)/3

let's find the value of (α + 2)/3 and (β + 2)/3 first

» (α + 2)/3 = (-4 + 2)/3 = -2/3

» (β + 2)/3 = (1 + 2)/3 = 3/3 or 1

sum of zeroes = -2/3 + 3/3

= 1/3

product of zeroes = -2/3 × 1

= -2/3

also -b/a = sum of zeroes and c/a = product of zeroes

therefore a = 3, b = -1 and c = -2

standard form of quadratic equation = ax² + bx + c

hence, the quadratic equation is 3x² - x - 2

option (e) none of the given option is correct =_=