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a positive real number is 8 less than another. if the sum of the squares of the two numbers is 80 find the numbers

User Corristo
by
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1 Answer

4 votes

Answer:

The numbers are


4+2√(6) and
-4+2√(6)

Explanation:

Let

x ------> one number

(x-8) ------> the another number

we know that


x^(2) +(x-8)^(2)=80


x^(2) +x^(2)-16x+64=80


2x^(2)-16x+64-80=0


2x^(2)-16x-16=0

Simplify


x^(2)-8x-8=0

The formula to solve a quadratic equation of the form
ax^(2) +bx+c=0 is equal to


x=\frac{-b(+/-)\sqrt{b^(2)-4ac}} {2a}

in this problem we have


x^(2)-8x-8=0

so


a=1\\b=-8\\c=-8

substitute in the formula


x=\frac{8(+/-)\sqrt{-8^(2)-4(1)(-8)}} {2(1)}


x=\frac{8(+/-)√(96)} {2}


x=\frac{8(+/-)4√(6)} {2}


x=4(+/-)2√(6)


x1=4(+)2√(6)


x2=4(-)2√(6)

so

the solution must be a positive real number


x=4(+)2√(6)


x-8=4(+)2√(6)-8 ------>
x-8=-4(+)2√(6)

User Zaher
by
8.1k points

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