Question

(01.03)

Solve the equation for x. If a solution is extraneous, be sure to identify it in your final answer.

square root of the quantity x minus 3 end quantity plus 5 equals x

Answer 1

= 7

Explanation:

Answer 2

x=7 satisfy the equation, so it is the solution.

x=4 doesn't satisfy the equation so it is extraneous solution.

Explanation:

The equation given is:

`√(x-3)+5=x`

Adding -5 on both sides

`√(x-3)=x-5`

Taking square on both sides

`(√(x-3))^2=(x-5)^2`

Now solving

`x-3 = x^2 -10x+25\\Arranging\\x^2-10x-x+3+25=0\\x^2-11x+28=0\\Factorizingx^2-7x-4x+28=0\\\\x(x-7)-4(x-7)=0\\(x-4)(x-7)=0\\x-4=0 \,\,and\,\, x-7=0\\x=4 \,\,and\,\, x=7`

Verifying solutions:

Putting x=4 in the equation

`√(x-3)+5=x`

`√(4-3)+5=4`

`√(1)+5=4`

`1+5=4`

`6\\eq 4`

So, x=4 doesn't satisfy the equation so it is extraneous solution.

Now Putting x=7 and verifying

`√(x-3)+5=x`

`√(7-3)+5=7`

`√(4)+5=7`

`2+5=7`

`7=7`

x=7 satisfy the equation, so it is the solution.