# What are the solutions of equation : square root (5x-4)+square root(11x+4)=4square rootx?

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### 2 Answers

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

x>=4/5

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.**