Question

quadratic equations...
class 10​

Answer 1

Given :----

• 6(x2 + 1/x2) – 25(x – 1/x) + 12 = 0

Solution :-------

Put (x – 1/x) = y,

squaring both sides now, we get,,

→ (x – 1/x)² = y²

⇒ x² + 1/x² – 2 = y² [ as (a-b)² = a² + b² -2ab ]

⇒ (x² + 1/x²) = (y² + 2)

Now, given equation becomes

6(y² + 2) – 25y + 12 = 0

⇒ 6y² + 12 – 25y + 12 = 0

⇒ 6y² – 25y + 24 = 0

⇒ 6y² - 16y - 9y + 24 = 0

⇒ 2y(3y – 8) – 3(3y – 8) = 0

⇒ (3y – 8)(2y – 3) = 0

⇒ 3y – 8 = 0 or 2y – 3 = 0

⇒ 3y = 8 or 2y = 3

⇒ 3y = 8 or 2y = 3

⇒ y = 8/3 or y = 3/2

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Putting both Equal to (x-1/x) now, and solving further ..

See solution in image now .....

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Hence, we can say that, value of x will be

[ 3, (-1/3) , 2 or (-1/2) ]

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#BAL

#anawerwithquality

Answer 2

Answer:

sorry dude I am not study in class 10th

....... I am in class 9th...

Step-by-step explanation:

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