If alpha ,beta are zeroes of the quadratic polynomial p (y) =3y^2-6y+4. Find the value of alpha by beta + beta by alpha +2×( one by alpha +one by beta)+ 3 alpha ×beta
Answers
Answer:
2\left(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\right)+2\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)+3\alpha\beta\\=8
Step-by-step explanation:
Given\: \alpha \:and\:\beta\\are\: zeroes\:of\:3x^{2}-6x+4
compare this with ax²+bx+c,we get
a = 3, b=-6, c=4,
i) Sum\:of \:the \: zeroes\\=\frac{-b}{a}\\=\frac{-(-6)}{3}\\=\frac{6}{3}\\\alpha+\beta=2 --(1)
ii) Product\:of\:the\: zeroes\\=\frac{c}{a}\\=\\alpha\beta=\frac{4}{3}---(2)
iii)\alpha^{2}+\beta^{2}\\=\left(\alpha+\beta\right)^{2}-2\alpha\beta\\=\left(2\right)^{2}-2\times \frac{4}{3}\\=4-\frac{8}{3}\\=\frac{12-8}{3}\\=\frac{4}{3}--(3)
Now,\\2\left(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\right)+2\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)+3\alpha\beta\\
=2\left(\frac{\alpha^{2}+\beta^{2}}{\alpha\beta}\right)+2\left(\frac{\beta+\alpha}{\alpha\beta}\right)+3\alpha\beta\\=2\left(\frac{\frac{4}{3}}{\frac{4}{3}}\right)+2\left(\frac{2}{\frac{4}{3}}\right)+3\times \frac{4}{3}
=2\times 1+2\times \frac{3}{2}+3
=2+3+3\\=8
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α/β + β/α +2×(1/α +1/β)+ 3 α×β = 9
Given:
α, β are the zeroes of the quadratic polynomial p(y) =3y²- 6y+4
To find:
The value of α/β + β/α +2×(1/α +1/β)+ 3 α×β
Solution:
Given p(y) = 3y²- 6y+4 and α, β are zeros
Compere given equation with ay²+by+c = 0
⇒ a = 3, b = -6 and c = 4
As we know sum of zero = -b/a
And product of zeros = c/a
⇒ α + β = -(-6)/3 = 2 ----(1)
⇒ α β = 4/3 ---(2)
Now find the value α/β + β/α +2×(1/α +1/β)+ 3 α×β using above values
α/β + β/α +2×(1/α +1/β)+ 3 α×β
⇒ (α²+β²)/αβ + 2(β+α)/αβ + 3αβ
From algebraic identity a²+b² = (a+b)²-2ab
⇒ (α+β)²- 2αβ /αβ + 2(β+α)/αβ + 3αβ
Now substitute α + β = 2 and αβ = 4/3 in above expression
⇒ (2)²- 2(4/3)/ 4/3 + 2(2)/ 4/3 + 3(4/3)
⇒ 4 - 8/3 / 4/3 + 4 / 4/3 + 4
⇒ 4/3 / 4/3 + 4 / 4/3 + 4
⇒ 4/3(3/4) + 4(3/4) + 4
⇒ 1 + 4 + 4 = 9
Therefore,
α/β + β/α +2×(1/α +1/β)+ 3 α×β = 9
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