Question

(01.03)

Answer 1

Answer:

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

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

Step-by-step explanation:

The equation given is:

$\sqrt{x-3}+5=x$

Adding -5 on both sides

$\sqrt{x-3}=x-5$

Taking square on both sides

$(\sqrt{x-3})^2=(x-5)^2$

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

$\sqrt{x-3}+5=x$

$\sqrt{4-3}+5=4$

$\sqrt{1}+5=4$

$1+5=4$

$6\neq 4$

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

Now Putting x=7 and verifying

$\sqrt{x-3}+5=x$

$\sqrt{7-3}+5=7$

$\sqrt{4}+5=7$

$2+5=7$

$7=7$

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