Math, asked by queensp73, 9 months ago

Hi friends !!
if alpha and beta are the zeroes of the polynomial p(x)=x²-5x+6,find the value of alpha⁴beta²+alpha²beta⁴​

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Answers

Answered by EliteSoul
102

Given:-

→ Polynomial : P(x) = x² - 5x + 6

To find:-

  • Value of αβ² + α²β

Solution:-

Here, p(x) = 0

∴ x² - 5x + 6 = 0

→ x² - (3 + 2)x + 6 = 0

→ x² - 3x - 2x + 6 = 0

→ x(x - 3) - 2(x - 3) = 0

→ (x - 3)(x - 2) = 0

At first,

→ (x - 2) = 0

→ x = 2

Again,

→ (x - 3) = 0

→ x = 3

Therefore,

Zeros of polynomial are: 3 & 2

\rule{200}{1}

Here,

  • α = 3
  • β = 2

Now we have to find value of :

→ α⁴β² + α²β⁴

Putting values we get:-

→ (3)⁴ × (2)² + (3)² × (2)⁴

→ (81 × 4) + (9 × 16)

→ 324 + 144

→ 468

Thus,

We get required value : 468 .

Answered by Cosmique
40

\underline{\huge{\bf{QuestioN}}}

If α and β are the zeroes of the polynomial p(x)=x² - 5x + 6 , find the value of α⁴β² + α²β⁴.

\underline{\huge{\bf{SolutioN}}}

We have,

p (x) = x² - 5x + 6

given that α and β are zeroes of p(x)

so,

\bf \alpha +\beta =\frac{-(coefficient\:of\:x)}{coefficient \:of\:x^2} \\\\\boxed{\bf \alpha +\beta =\frac{-(-5)}{1} = 5}.........eqn(1)

also,

\bf \alpha \beta = \frac{constant \:term}{coefficient\:of\:x^2} \\\\\boxed{\bf \alpha \beta = \frac{6}{1} =6}..........eqn(2)

we have to find;

→α⁴β² + α²β⁴

taking α²β² common

→α²β² ( α² + β² )

using identity x²+y² = (x+y)²-2xy

→(αβ)² ( (α+β)²- 2αβ)

putting values of α+β & αβ using eqn(1) & (2)

→(6)² ( ( 5)² - 2 ( 6 ) )

→ 36 ( 25 - 12 )

→ 36 (13)

→ 468  (Ans.)

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