Question 390943
Find a number such that one half of it's square is equal to twice the number plus 8?
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Let number be x
1/2 *(x^2)=2x+8
multiply by 2
x^2=2(2x+8)
x^2=4x+16
x^2-4x-16=0
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use quadratic formula to find the roots of the equation
{{{x=(-b+-sqrt(b^2-4ac))/2a}}}
a=1,b=-4,c=16
b^2-4ac=80
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{{{x1=(-b+sqrt(80))/2}}}
{{{(highlight(x1=6.47))}}}
{{{x2=(-b-sqrt(80))/2}}}
{{{(highlight(x2=-2.47))}}}
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m.ananth@hotmail.ca