Question

If x+2y=10 and xy=15, find the value of x^2 +4y^2

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Answer 1

$\huge\boxed{\fcolorbox{blue}{blue}{question-:}}$

If x+2y=10 and xy=15, find the value of x^2 +4y^2

$\huge\boxed{\fcolorbox{blue}{blue}{solution-:}}$

To find= x^2 +4y^2

Squaring both sides (x+2y)^2=(10)^2

(x)^2 + (2y)^2 +2 × x × 2y =100

x^2 +4y^2 +4xy= 100

x^2 +4y^2+4×15 = 100

x^2 + 4y^2 + 60 =100

x^2 +4y^2= 100-60 = 40

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Answer 2

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$\large{\underbrace{\sf{\purple{required \: solution:}}}}$

$\longrightarrow \sf 40$

Given that,

• x + 2y = 10
• xy = 15

$\sf Equation \rightarrow {x}^{2} + {4y}^{2}$

On squaring x + 2y , we have :

$\rightarrow \sf {(x + 2y)}^{2} = 100$

$\rightarrow \sf {x}^{2} + 4xy + 4 {y}^{2} = 100$

$\rightarrow \sf {x}^{2} + 4 {y}^{2} = 100 - 4xy$

$\rightarrow \sf {x}^{2} + 4 {y}^{2} = 100 - 4(15)$

• Given that xy = 15

$\rightarrow \sf {x}^{2} + 4 {y}^{2} = 100 -60$

$\rightarrow \boxed{ \purple {\sf {x}^{2} + 4 {y}^{2} = 40}}$

$\large{\underbrace{\sf{\purple{ Know \: more : }}}}$

• The first term is the product of two binomials give third terms.
• The middle term = ( 1st term of 1st binomial × 2nd term of 2nd binomial ) + ( 2nd term of 1st binomial × 1st term of 2nd binomial ) = Product of outer terms × product of inner terms.
• The third term = Product of 2nd terms of the two binomials.
• Use these identities while solving expressions:
• (a + b)² = a²+ 2ab + b²
• (a - b)²= a² - 2ab + b²
• (a + b) (a - b) = a² - b²

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