Question

if x+2y=9 and xy=13, find the value of x sq. +. 4y sq.​

Answer 1

$\huge\boxed{\red{\mathtt{Question}}}$

Given :- x+2y=9

xy =13

To find $\sf x{}^{2}+4y{}^{2}$

$\huge\mathbb{\underline{SOLUTION}}$

$\large\mathbb{\underline{\red{FORMULA\:USED}}}$

$\large\sf (a+b){}^{2} =a{}^{2}+b{}^{2}+2ab$

$\large\bf\underline{According \:to\: question}$

$\sf x{}^{2}+4y{}^{2}= (x+2y){}^{2}-4xy\\ \sf\implies x{}^{2}+4y{}^{2}= x{}^{2}+4y{}^{2}+4xy-4xy\\ \sf\implies x{}^{2}+4y{}^{2}=x{}^{2}+4y{}^{2}\\ \sf\implies x{}^{2}+4y{}^{2}= (9){}^{2} -4×13\\ \sf\implies x{}^{2}+4y{}^{2}= 81- 52\\ \sf\implies x{}^{2}+4y{}^{2}=29$

$\large{\underline{\boxed{\red{\bf{x{}^{2}+4y{}^{2}=29}}}}}$

$\large\bf\underline{Some\: important\: formula}$

$\large\sf1)(a+b){}^{2}=a{}^{2}+b{}^{2}+2ab\\ \sf\large 2) (a-b){}^{2}=a{}^{2}+b{}^{2}-2ab\\ \sf\large 3) (a+b+c){}^{2}=a{}^{2}+b{}^{2}+c{}^{2}+2ab+2bc+2ca\\ \sf\large 4) (a-b){}^{3}=a{}^{3}+b{}^{3}+3ab(a+b)\\ \sf\large 5) (a-b){}^{3}=a{}^{3}-b{}^{3}-3ab(a+b)\\ \sf\large 6) a{}^{3}+b{}^{3}=(a+b)(a{}^{2}+b{}^{2}-ab)\\ \sf\large 7) a{}^{3}-b{}^{3}=(a-b)(a{}^{2}+b{}^{2}+ab)\\ \sf\large 8) a{}^{3}+b{}^{3}+c{}^{3}-3abc=(a+b+c)(a{}^{2}+b{}^{2}+c{}^{2}-ab-bc-ca)$