find the difference between two positive numbers is 4 and the sum of their squares is 58. Find the numbers

the bigger number is=

the smaller number is =

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a-b=4

a^2+b^2=58

a-b=4 implies a=b+4

substitute

(b+4)^2+b^2=58

b^2+8b+16+b^2=58

2b^2+8b+16=58

2b^2+8b+16-58=0

2b^2+8b-42=0

b^2+4b-21=0

(b+7)(b-3)=0

b-3=0

b=3

a-3=4

a=7

a=7,b=3

don't forget, you said two positive numbers so we stop here

Michael F. | Mathematics TutorMathematics Tutor

A-B=4 (1)

A²+B²=58 (2)

(A-B)²=A²-2AB+B²=16

Substitute from equation (2)

(A-B)²=-2AB+58=16

So now have -2AB+58=16 or that -2AB=-42 or that AB=21

Look at A-B=4 and AB=21

Substitute A=B+4 into AB=21 and have (B+4)B=21 or that

B²+4B-21=0 which factors int (B+7)(B-3)=0 which gives that B=3 or B=-7

For B=3, A=7 and A²+B²=58

For B=-7, A=-11 and A²+B²=170 not a solution

Smaller number =3; larger number =7

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