Question 76673
Square root of 2t minus t + 4 = 0
Is this what you meant?
{{{sqrt(2t)-t+4=0}}}
First isolate the radical
{{{sqrt(2t)-t+t+4-4=0+t-4}}}
{{{sqrt(2t)=t-4}}}
Now square both sides
{{{(sqrt(2t))^2=(t-4)^2}}}
{{{2t=(t-4)(t-4)}}}
{{{2t=t^2-4t-4t+16}}}
{{{2t=t^2-8t+16}}}
Set the problem = to 0
{{{2t-2t=t^2-8t-2t+16}}}
Factor
{{{0=t^2-10t+16}}}
{{{0=(t-8)(t-2)}}}
Set each factor = to 0 and solve for x
{{{t-8=0}}} or {{{t-2=0}}}
{{{t-8+8=0+8}}} or {{{t-2+2=0+2}}}
{{{t=8}}} or {{{t=2}}}
Check for extraneous solutions:
{{{sqrt(2(8))-(8)+4=0}}} or {{{sqrt(2(2))-(2)+4=0}}}
{{{sqrt(16)-8+4=0}}} or {{{sqrt(4)-2+4=0}}}
{{{4-8+4=0}}} or {{{2-2+4=0}}}
{{{0=0}}} or {{{0=0}}}
Both solutions work!
{{{highlight(t=8)}}} or {{{highlight(t=2)}}}
Happy Calculating!!!!