# Given the equation x - 4 =√2x, solve for x and identify if it is an extraneous solution.

**Solution:**

x - 4 = √2x --- (1)

Squaring both sides we get

(x - 4)^{2} = (√2x)^{2}

x^{2} - 8x + 16 = 2x^{2}

x^{2 }+ 8x - 16 = 0

x = -8 ± √8^{2} - 4(1)(-16) / 2(1)

x = -8 ± √64 + 64) / 2

x = -8 ± √128 / 2

x = -4 ± 8/√2

The two roots of the quadratic equation are - 4 + 8/√2 and -4 - 8/√2.

These also can be written as a) -4 + 4√2 and b) -4 - 4√2

Let us substitute a) i.e. (-4 + 4√2) in equation 1 and we have

(-4 +4√2 - 4) = √2(-4 + 4√2)

-8 + 4√2 = 8 - 4√2

Now the above is incorrect.LHS ≠ RHS, Hence ( -4 + 4√2) is an extraneous solution.

Let us substitute b) i.e. -4 - 4√2 in equation 1) and we get,

-4 - 4√2 - 4 = √2(-4 - 4√2)

-8 - 4√2 = -8 - 4√2

Hence both sides are indeed equal and hence the solution to the equation is -8 - 4√2.

## Given the equation x - 4 =√2x, solve for x and identify if it is an extraneous solution.

**Summary:**

The given equation x - 4 =√2x, when solved for x gives one solution -8 - 4√2. and other solution -8 + 4√2 is an extraneous solution.

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