Question

determine 5 is root of equation √x^2-4x-5 +√x^2-25=√5x^2-24x-5​

Answer 1

Step-by-step explanation:

Given :-

√(x²-4x-5)+√(x²-25) =√(5x²-24x-5)

To find:-

Determine 5 is the root of the equation ?

Solution :-

Given equation is

√(x²-4x-5)+√(x²-25) =√(5x²-24x-5)

We know that

If a number is a root of the given equation then it satisfies the given equation.

Put x = 5 in LHS then

=> √[5²-4(5)-5]+√(5²-25)

=> √(25-20-5)+√(25-25)

=> √(25-25)+√0

=> √0+√0

=> 0+0

=> 0

And Put x = 5 in RHS

√[5(5²)-24(5)-5]

=> √[5(25)-120-5]

=> √(125-125)

=> √0

=> 0

=> LHS = RHS is true for x = 5

So, 5 is the root of the equation.

Answer:-

5 is the root of the given equation.

Used Concept:-

→ Let k be any real number then on putting k in the place of the variable then LHS = RHS then K be the root or solution of the equation .

→ If a number is a root of the given equation then it satisfies the given equation.

Answer 2

$\boxed{ answer }$

If a number is a root of the given equation then it satisfies the given equation.

Put x = 5 in LHS then

=> √[5²-4(5)-5]+√(5²-25)

=> √(25-20-5)+√(25-25)

=> √(25-25)+√O

=> √O+√O

=> 0+0

=> 0

And Put x = 5 in RHS

√[5(5²)-24(5)-5]

=> √[5(25)-120-5]

=> √(125-125)

=> √O

=> 0

=> LHS = RHS is true for x = 5

So, 5 is the root of the equation.

5 is the root of the given equation.