Need help. Need someone to walk through with solving this surd equation: 3 sqrtx = x-10
\[3\sqrt{x}=x-10\]
when u square both sides u will get a quadratic equation \[x ^{2}-29x+100\] now solve this equation and substitute the values of x in ur equation and see which value satisfies it
so i got x = 25 and x = 4 but im confused on how to substitute the values of x
therefore for x=25 \[3\sqrt{x}\]=3sqrt25=3(5)=15 and rhs=25-10=15 so rhs=lhs so it satisfies the equation now substitute at x=4 and see if lhs = rhs
arent you meant to substitute in x =25 for the whole equation of 3 sqrtx = x-10 not just 3sqrt x?
oh dw you did, didnt read it properly
yes you have to substitute the value of x in both sides and see if they give the same value .
so is x =25 the only solution as x =4 i got 6 and - 6
yes so x=25 is ur solution
thank you very much for your help
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