if alpha and beta are two zeros of the 2 x square - 5 x minus 7 then find the polynomial whose zeros are 2 alpha + 3 Beta and 3 alpha + 2 Beta
Answers
2x²-5x-7
2x²-7x+2x-7
x(x-7) 1(x-7)
so x=7
x=-1
alpha=μ=7
β=-1
another equation zeroes are 2μ+3β,3μ+2β
2(7)+3(-1),3(7)+2(-1)
11,19
equation of the given zeroes
x²+(μ+β)x-μβ
x²+(11+19)x-11*19
x²+30x-209
I HOPE THIS WILL HELP U
Answer:
Step-by-step explanation:
HI !
NOTE :-
α² + β² can be written as (α + β)² - 2αβ
p(x) = 2x² - 5x + 7
a = 2 , b = - 5 , c = 7
α and β are the zeros of p(x)
we know that ,
sum of zeros = α + β
= -b/a
= 5/2
product of zeros = c/a
= 7/2
===============================================
2α + 3β and 3α + 2β are zeros of a polynomial.
sum of zeros = 2α + 3β+ 3α + 2β
= 5α + 5β
= 5 [ α + β]
= 5 × 5/2
= 25/2
product of zeros = (2α + 3β)(3α + 2β)
= 2α [ 3α + 2β] + 3β [3α + 2β]
= 6α² + 4αβ + 9αβ + 6β²
= 6α² + 13αβ + 6β²
= 6 [ α² + β² ] + 13αβ
= 6 [ (α + β)² - 2αβ ] + 13αβ
= 6 [ ( 5/2)² - 2 × 7/2 ] + 13× 7/2
= 6 [ 25/4 - 7 ] + 91/2
= 6 [ 25/4 - 28/4 ] + 91/2
= 6 [ -3/4 ] + 91/2
= -18/4 + 91/2
= -9/2 + 91/2
= 82/2
= 41
-18/4 = -9/2 [ simplest form ]
a quadratic polynomial is given by :-
k { x² - (sum of zeros)x + (product of zeros) }
k {x² - 5/2x + 41}
k = 2
2 {x² - 5/2x + 41 ]
2x² - 5x + 82 -----> is the required polynomial