Question 623062
Please put radicands in parentheses. ("Radicand" is the name for the expression inside a radical.) The way you posted the problem it is not possible to know if it is:
{{{7x-37sqrt(x+48) = 0}}}
or
{{{7x-37sqrt(x)+48 = 0}}}<br>
Since I don't know which one is real, I'm just going to get you started on both and tell you how to finish them.<br>
{{{7x-37sqrt(x+48) = 0}}}
Isolate the square root by adding it to both sides:
{{{7x = 37sqrt(x+48)}}}
Then<ol><li>Square both sides. This will eliminate the square root.</li><li>Solve the equation. (It will be a quadratic equation.)</li><li>Check each solution. This is <i>not optional</i> since you've squared both sides of the equation. Use the original equation to check and reject any "solutions" that do not check out.</li></ol>
{{{7x-37sqrt(x)+48 = 0}}}
This one can be solved in a similar way to the one above. But there is a faster way. Since the exponent on x in 7x is 1 and the exponent on x in {{{37sqrt(x)}}} is 1/2 (Remember that square roots can be expressed as exponents of 1/2?) Since the exponent on x is twice as large as the exponent on {{{sqrt(x)}}}, this equation is in what is called quadratic form. It is not literally a quadratic equation but it has the same pattern of exponents. And we can solve this in a similar way.<br>
<i>Exactly</i> like
{{{7q^2-37q+48=0}}}
can be factored into
{{{(7q-16)(q-3) = 0}}}
{{{7x-37sqrt(x)+48 = 0}}}
can be factored into
{{{(7sqrt(x)-16)(sqrt(x)-3) = 0}}}
We can now use the Zero Product Property which tells us that one of these factors must be zero:
{{{7sqrt(x)-16 = 0}}} or {{{sqrt(x)-3 = 0}}}
Solving these...
{{{7sqrt(x) = 16}}}
{{{sqrt(x) = 16/7}}}
{{{(sqrt(x))^2 = (16/7)^2}}}
{{{x = 256/49}}}<br>
{{{sqrt(x)-3 = 0}}}
{{{sqrt(x)= 3}}}
{{{(sqrt(x))^2= (3)^2}}}
{{{x = 9}}}<br>
Just like the other problem, we <i>must</i> check our answers because we squared both sides of the equation. I'll leave that up to you.