Question 155598
Remember: For rectangle,
{{{Area=Length*Width}}}----------------------------> working eqn
Given:
{{{L=unknown=x}}}
{{{W=x-2ft}}}
{{{A=8ft^2}}}
Substitute, {{{8=x*(x-2)}}}
{{{8=x^2-2x}}} ------> {{{x^2-2x-8=0}}}
BY QUADRATIC: {{{a=1}}}, {{{b=-2}}}, & {{{c=-8}}}
{{{x=(-b+-sqrt(b^2-4ac))/(2a)}}}
{{{x=(-(-2)+-sqrt(-2^2-4*1*-8))/(2*1)}}}
{{{x=(+2+-sqrt(4+32))/2}}}
{{{x=(2+-sqrt(36))/2}}}
{{{x=(2+-6)/2}}}---> 2 values {{{(2+6)/2=8/2=4}}},{{{(2-6)/2=-4/2=-2}}}
USED {{{x=4ft}}} ----------------> LENGTH
{{{4-2=2ft}}} -----------------> WIDTH
In doubt? Go back working eqn:
{{{8ft^2=(4ft)(2ft)}}}
{{{8ft^2=8ft^2}}}
Thank you,
Jojo