Question

the difference between two positive numbers is 7 and the square of their sum is 289. find the two numbers

Answer 1

please mark it as brainliest

Answer 2

Answer:

$\red { Two \: numbers \: are } \green { \: 15 \:and \: 8}$

Step-by-step explanation:

$Let \: x \: and \: (x-7) \: are \: two \: positive \\numbers$

/* According to the problem given

$Sum \: of \: the \: squares \: two \: numbers =289$

$\implies x^{2} + (x-7)^{2} = 289$

$\implies x^{2} + x^{2} - 2\times x \times 7 + 7^{2} - 289 = 0$

$\implies 2x^{2}-14x+ 49-289=0$

$\implies 2x^{2} - 14x - 240 = 0$

/* Divide each term by 2, we get

$\implies x^{2} - 7x - 120 = 0$

/* Splitting the middle term, we get

$\implies x^{2} -15x + 8x - 120 = 0$

$\implies x( x - 15 ) + 8( x - 15 ) = 0$

$\implies (x -15)( x+8) = 0$

$\implies x - 15 = 0 \: Or \: x + 8 = 0$

$\implies x = 15 \: Or \: x = -8$

/* Numbers are positive .

Therefore.,

$Two \: numbers \:are \: x = 15\:and L\\x-7 = 15-7 = 8$

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