Question

79) If α and β are the roots of x² -2x + 3 = 0, then the
equation with roots α+2, β+2 is
A) x² + 6x - 11 = 0
B) x² -11x + 6 = 0
C) x² - 6x + 11 = 0
D) x² +11x - 6 = 0

Answer 1

Answer:

c)

${x}^{2} - \: 6x \: + 11$

Answer 2

Answer:

answer is C

alpha + beta = -b/a

=-(-2)/1

=2.....(i)

now,

sum of the zeros of required poly.= a+b

=(a+2)+(b+2)

from (i)

=a+b+4

=2+4

=6

also, product(ab)=d/a

=3/1

=3

product of zeros of other one=

(a+2)(b+2)

=ab+2a+2b+4

=3+4+2(a+b)

=7+2(2)

=7+4

=11

therefore,

x^2-( sum of zeros)x+(product)

=x^2-6x+11