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if the sum of two numbers is four and the sum of their squares minus three times their product is 76, find the numbers

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Answer:

Numbers =(6,-2 )

Step-by step explanation:

Given

the sum of two numbers is four and the sum of their squares minus three times their product is 76,

Let the first term =X

Second term will be=4-x

So

X^2+(4-x)^2-3x(4-x)=76

x^2+16+x^2-8x-12x+3x^2=76

5x^2-20x+16=76

5x^2-20x-60=0

5(x^2-4x-12)=0

So x^2-4x-12=0

x^2-(6-2)x-12=0

x^2-6x+2x-12=0

X(x-6)+2(x-6)=0

(X-6)(X+2)=0

X=6,-2

Let X=6 and -2 then numbers will be 6 and (4-6)=-2

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