What are the solutions of equation : square root (5x-4)+square root(11x+4)=4square rootx?
We have to solve sqrt (5x-4) + sqrt(11x+4) = 4*sqrt x
sqrt (5x-4) + sqrt(11x+4) = 4*sqrt x
square both the sides
=> (5x - 4) + (11x + 4) + 2*sqrt (5x-4)*sqrt(11x+4) = 16x
=> 16x + 2*sqrt (5x-4)*sqrt(11x+4) = 16x
=> 2*sqrt (5x-4)*sqrt(11x+4) = 0
=> (5x - 4)(11x + 4) = 0
5x - 4 = 0 , gives x = 4/5
11x + 4 = 0, gives x = -4/11. But sqrt (-4/11) is not real.
Therefore the real value of x is 4/5.
Before solving the equation, we'll impose constraints of existence of the radicals.
We'll impose the condition that each radicand to be positive:
5x - 4 >= 0
5x >= 4
11x + 4 >= 0
11x >= -4
x >= -4/11
x >= 0
The common range of admissible values for x is [4/5 ; +infinite).
We'll solve the equation by raising to square both sides:
5x - 4 + 11x + 4 + 2sqrt(5x-4)(11x+4) = 16x
We'll combine and eliminate like terms:
16x + 2sqrt(5x-4)(11x+4) = 16x
We'll subtract 16x:
2sqrt(5x-4)(11x+4) = 0
We'll divide by 2 and we'll raise to square to eliminate the radical:
(5x-4)(11x+4) = 0
We'll set the 1st factor equal to 0.
5x - 4 = 0
5x = 4
x = 4/5
We'll set the 2nd factor as zero:
11x + 4 = 0
x = -4/11
Since the value x = -4/11 doesn't belong to the interval of admissible values, we'll reject it.
The solution of the equation is x = 4/5.