Question

have a question about quadratic equations​

Answer 1

Step-by-step explanation:

1 . The sum of squares of 2 consecutive numbers is 85 .

Let the two numbers be x and x+1.

${x}^{2} + {(x + 1)}^{2} = 85 \\ {x}^{2} + {x }^{2} + 2(x)(1) + {1}^{2} = 85 \\ 2 {x}^{2} + 2x + 1 = 85 \\ 2 {x}^{2} + 2x - 84 = 0 \\ dividing \: the \: equation \: by \: 2 \\ {x}^{2} + x - 42 = 0 \\ {x}^{2} + 7x - 6x - 42 = 0 \\ x(x + 7) - 6(x + 7) = 0 \\ (x - 6)(x + 7) = 0 \\ x = 6 \: (or) \: x = - 7$

THEREFORE THE TWO CONSECUTIVE INTEGERS WHOSE SUM OF SQUARES IS 85 IS 6 AND 7 (or) -7 AND -6 !

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Let the 2 parts of 29 be x and y .

$x + y = 29 \: \: \: \: (1) \\ {x}^{2} + {y}^{2} = 425 \: \: \: \: (2) \\ from(1) \: y =29 - x \\ substitute \: y = 29 - x \: in(2) \\ {x}^{2} + {(29 - x)}^{2} = 425 \\ {x}^{2} + {29}^{2} - 2(29)(x) + {x}^{2} = 425 \\ {x}^{2} + {x}^{2} - 58x + 841 = 425 \\ 2 {x}^{2} - 58x + 416 = 0 \\ divide \: the \: equation \: by \: 2 \\ {x}^{2} - 29x + 208 = 0 \\ {x}^{2} - 13x - 16x + 208 = 0 \\ x(x - 13) - 16(x - 13) = 0 \\ (x - 16)( x - 13) = 0 \\ x = 16 \: (or) \: x = 13 \\ if \: x = 16 \: \: \: y = 29 - 16 = 13 \\ \: \: \: \: \: x = 13 \: \: \: y = 29 - 13 = 16$

THEREFORE THE PARTS OF 29 WITH THE SUM OF SQUARES IS 425 IS 13 AND 16 !

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3 . The third question is not clear and full !

So I can't answer the 3rd question !

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I HOPE IT IS HELPFUL !

HAVE A NICE DAY WITH BRAINLY !