solve the 19 th one plss
Answers
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
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 =_=