Question

If (X^2 +Y^2)/XY = 170/13 and x = 3, then y = ?
A. 24
B. 27
C. 39
D. 51
E. 219

Answer 1

from given we know

(x^2+y^2)/xy = 170/13

and

x=3

substituting x = 3 in given equation

(9+y^2)/3y = 170/13

13y^2-510 y +117=0

y={510+or - √(260100-6084)}/26

y={510+ or - √254016}/26

y={510 + or - 504}/26

y=(510-504)/26 or (510+504)/26

y=6/26 or 1014/26

y=6/26 or 39

therefore the value of y=39

Answer 2

Answer:

If a and b are the roots of the equation x^2 +x+2=0, then ( a^10 + b^10) / [a^-10 + b^-10)

A 4096

B. 2048

C. 1024

D. 512

E. 256

\huge\underline{\underline{\texttt{{Solution:}}}}

Solution:

Option C. 1024

\huge\underline{\underline{\texttt{{Explanation:}}}}

Explanation:

According to the question

Here α and β are the roots of the equation,

⇒ x² + x + 2 = 0.

Therefore,

⇒α + β = -1/1

or,

⇒ α + β = -b/a.

⇒ α + β = -1.

Products of zeroes of quadratic equation,

⇒ αβ = c/a.

⇒ αβ = 2.

Now,

We have to find the value of

(α¹⁰+β¹⁰)/{α^(-10)+β^(-10)}

=(α¹⁰+β¹⁰)/(1/α¹⁰ +1/β¹⁰)

={(α¹⁰+β¹⁰)/(α¹⁰+β¹⁰)}×(αβ)¹⁰

=(αβ)¹⁰

=(2)¹⁰

=1024

Hence, option C is correct

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this is answer for your upper question.