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Alja [10]
3 years ago
9

(01.03)

Mathematics
1 answer:
Fittoniya [83]3 years ago
3 0

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.

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