# Check whether x = 4 and x = 5 are solutions of the given equation :√(x2 – 16) – (x – 4) = √(x2 – 5x + 4).

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### 4 Answers

To check whether x=4 and x=5 are solutions to the equation

√(x2 – 16) – (x – 4) = √(x2 – 5x + 4).

Solution:

When x=4 , LHS sqrt (4^2--16)-(4-4) = 0. RHS: sqrt(4^2-5*4+4) = sqrt(16-20+4)=0. So LHS = RHS when x=4.So the equation is satisfied when x= 4. So x is asolution.

When x=5:

LHS = sqrt(5^2-16)-(5-4) = 3-1 =2

RHS: sqrt(5^2-5*5+4) =2. So LHS = RHS when x=5. Therefore, 5 is aroot of the equation.

we can solve it,

sqr root{(x^2-16)-(x-4)}=(x^2-5x+4)^1/2

or(x^2-16-x+4)^1/2=(x^2-5x+4)^1/2

or,(x^2-x-12)^1/2=(x^2-5x+4)^1/2

or,(x^2-x-12)=(x^2-5x+4)

or,4x=16

so,x=4

this equation have a unique solution.means x=4

so x=5 is not satisfies for this equation.

We can check if a given value of a variable is a solution by substituting the value of variable in the equation and checking if the equation holds true.

Thus to check if x = 4 and x = 5 solution of the equation:

(x^2 - 16)^1/2 - (x - 4) = (x^2 - 5x + 4)^1/2 we first substitute the value x = 4 and check if the equation is valid. Then we repeat the process with the value x = 5.

**Checking if x = 4 is a solution:**

(4^2 - 16)^1/2 - (4 - 4) = (4^2 - 5*4 + 4)^1/2

Simplifying this we get:

(16 - 16)^1/2 - 0 = (16 - 20 + 4)^1/2

0^1/2 = 0^1/2

As left hand side of this equation is equal to right hand side, x = 4 is a solution for the given equation.

**Checking if x = 5 is a solution:**

(5^2 - 16)^1/2 - (5 - 4) = (5^2 - 5*5 + 4)^1/2

Simplifying this we get:

(25 - 16)^1/2 - 1 = (25 - 25 + 4)^1/2

9^1/2 - 1 = 4^1/2

3 - 1 = 2

2 = 2

As left hand side of this equation is equal to right hand side, x = 5 is a solution for the given equation.

PS:

Answers in the next two posts are different because of difference in interpreting the given equation.

While I have assumed that square root sigh on the left hand side of the equation applies only to the terms (x^2 - 16), the other two answer apply this sign to all the terms on the left hand side f equation.

By the way if we accept that the square root sign is applicable to all the terms on the left hand side of the equation then the equation can be simplified by removing the square root sign from both the sides.

I'll verify if the values given are the solution of the equation, considering the equation written in the following form:

sqrt[(x^2-16)-(x-4)]=sqrt(x^2-5x+4)

It is very simple to find out if the 2 given values are the given solution for the equation:

sqrt[(x^2-16)-(x-4)]=sqrt(x^2-5x+4)

First of all, we'll try to substitute the value x = 4 into the equation, to see if the value verifies it.

sqrt[(4^2-16)-(4-4)]=sqrt(4^2-5*4+4)

sqrt[(16-16)-0]=sqrt(16-20+4)

sqrt 0=sqrt 0

0=0, so **x=4** is the value for the equation above.

Now, we'll try to substitute the value x = 5 into the equation, to see if the value verifies it.

sqrt[(5^2-16)-(5-4)]=sqrt(5^2-5*5+4)

sqrt(25-16-1)=sqrt(25-25+4)

sqrt 8=sqrt 4

2sqrt 2=2, NOT TRUE!

So x=5 is not the root for the given equation.