Solve for x
x=7
x=4
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\sqrt{333+49x}=19+x
Subtract -x from both sides of the equation.
\left(\sqrt{333+49x}\right)^{2}=\left(19+x\right)^{2}
Square both sides of the equation.
333+49x=\left(19+x\right)^{2}
Calculate \sqrt{333+49x} to the power of 2 and get 333+49x.
333+49x=361+38x+x^{2}
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(19+x\right)^{2}.
333+49x-361=38x+x^{2}
Subtract 361 from both sides.
-28+49x=38x+x^{2}
Subtract 361 from 333 to get -28.
-28+49x-38x=x^{2}
Subtract 38x from both sides.
-28+11x=x^{2}
Combine 49x and -38x to get 11x.
-28+11x-x^{2}=0
Subtract x^{2} from both sides.
-x^{2}+11x-28=0
Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
a+b=11 ab=-\left(-28\right)=28
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-28. To find a and b, set up a system to be solved.
1,28 2,14 4,7
Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 28.
1+28=29 2+14=16 4+7=11
Calculate the sum for each pair.
a=7 b=4
The solution is the pair that gives sum 11.
\left(-x^{2}+7x\right)+\left(4x-28\right)
Rewrite -x^{2}+11x-28 as \left(-x^{2}+7x\right)+\left(4x-28\right).
-x\left(x-7\right)+4\left(x-7\right)
Factor out -x in the first and 4 in the second group.
\left(x-7\right)\left(-x+4\right)
Factor out common term x-7 by using distributive property.
x=7 x=4
To find equation solutions, solve x-7=0 and -x+4=0.
\sqrt{333+49\times 7}-7=19
Substitute 7 for x in the equation \sqrt{333+49x}-x=19.
19=19
Simplify. The value x=7 satisfies the equation.
\sqrt{333+49\times 4}-4=19
Substitute 4 for x in the equation \sqrt{333+49x}-x=19.
19=19
Simplify. The value x=4 satisfies the equation.
x=7 x=4
List all solutions of \sqrt{49x+333}=x+19.
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