if aplha and beta are zeros of polynomial2x^2-5x+7 find the polynomial whose zeros are 2alpha+3beta and 3alpha +2beta
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2x^2-5x+7
alpha + beta =-b/a=-(-5)/2=5/2
alpha * beta=c/a=7/2
By formula ,( taking alpha as a and beta as b)
x^2-(sum of zeroes )x+(product of zeroes)
x^2-(2a+3b+3a+2b)x+{(2a+3b)(3a+2b)}
x^2-(4a+4b)x+(6a^2+4ab+9ab+6b^2)
x^2-[4(a+b)]x+[6(a^2b^2)+13ab]
now,
a+b=5/2 and ab=7/2
x^2-[4*5/2]x+[6(7/2)^2+13*7/2
x^2-(10)x+(147/2+91/2)
x^2-10x+238/2
x^2-10x+119 is the required polynomial.
alpha + beta =-b/a=-(-5)/2=5/2
alpha * beta=c/a=7/2
By formula ,( taking alpha as a and beta as b)
x^2-(sum of zeroes )x+(product of zeroes)
x^2-(2a+3b+3a+2b)x+{(2a+3b)(3a+2b)}
x^2-(4a+4b)x+(6a^2+4ab+9ab+6b^2)
x^2-[4(a+b)]x+[6(a^2b^2)+13ab]
now,
a+b=5/2 and ab=7/2
x^2-[4*5/2]x+[6(7/2)^2+13*7/2
x^2-(10)x+(147/2+91/2)
x^2-10x+238/2
x^2-10x+119 is the required polynomial.
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