Question

$[tex]\bf{if \: x+y=1,xy(xy-2)=12, x^4+y^4=?}$[/tex]
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Answer 1

Gɪᴠᴇɴ :-

• x + y = 1
• xy(xy - 2) = 12

Tᴏ Fɪɴᴅ :-

• x⁴ + y⁴ = ?

Sᴏʟᴜᴛɪᴏɴ :-

→ (x + y) = 1

Squaring Both Sides we get,

→ (x + y)² = (1)²

using (a + b)² = a² + b² + 2ab in LHS , we get,

→ x² + y² + 2xy = 1

→ x² + y² = (1 - 2xy) ------------- Equation ❶

____________

Now,

➻ x⁴ + y⁴

Adding & Subtracting 2x²y² we get,

➻ x⁴ + y⁴ + 2x²y² - 2x²y²

➻ (x⁴ + y⁴ + 2x²y²) - 2x²y²

Comparing it with a² + b² + 2ab = (a + b)² Now,

➻ (x² + y²)² - 2x²y²

Putting value of Equation ❶ Now,

➻ (1 - 2xy)² - 2x²y²

using (a - b)² = a² + b² - 2ab Now, we get

➻ 1 + (2xy)² - 2*1*2xy - 2x²y²

➻ 1 + 4x²y² - 4xy - 2x²y²

➻ 1 - 4xy + 4x²y² - 2x²y²

➻ 1 - 4xy + 2x²y²

➻ 1 + 2x²y² - 4xy

➻ 1 + 2xy(xy - 2)

Finally, Putting given value of xy(xy - 2) = 12 Now, we get,

➻ 1 + 2*12

➻ 1 + 24

➻ 25 (Ans.)

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For One More Solution Refer To Image Now.

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Answer 2

$\rule{200}2$

$\huge\tt{GIVEN:}$

• X + Y = 1,
• xy(xy - 2) = 12

$\rule{200}2$

$\huge\tt{TO~FIND:}$

• x⁴ + y⁴ = ?

$\rule{200}2$

$\huge\tt{SOLUTION:}$

↪( x+y ) = 1

↪( x+y )² = (1)² (Squaring both sides)

↪x² + y² + 2xy = 1(using (a+b)² = a² + b² + 2ab in LHS)

↪x² + y² + 2xy = 1

↪x² + y² = (1 - 2xy) ___(EQ.1)

So,

↪x⁴ + y⁴ (Adding & Subtracting 2xy²y²)

↪x⁴ + y⁴ + 2xy²y² - 2xy²y²

↪ (x⁴ + y⁴ + 2xy²y²) - 2xy²y²

↪(x² + y²)² - 2xy²y² (comparing it with a² + b² + 2ab= a +b²)

↪(1 - 2xy)² - 2xy²y²

↪1 + (2xy)² - 2 × 1 × 2xy - 2xy²y² (putting the values of equation.1 )

↪1 + 4xy²y² - 4xy - 2x²y² (using (a-b)² = a² + b² - 2ab )

↪1 + 2xy²y² - 4xy

↪1 + 2xy(xy - 2)

↪1 + 12 × 2 (using the value of xy(xy-2) = 12)

↪1 + 24

↪25

$\rule{200}2$