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# Given the equation √2x - 5 = 2, solve for x and identify if it is an extraneous solution.

**Solution:**

The given equation is: x√2 - 5 = 2

x√2 = 2 + 5

x√2 = 7

Squaring both sides we have

2x^{2} = 49

x^{2} = 49/2

x = ±7/√2

To test whether there is an extraneous solution or not we will substitute the two values in the original equation and verify.

1. x = +7/√2

Substituting x = +7/√2 in (x√2 - 5 = 2) we get,

(7/√2)√2 - 5 = 2

7 - 5 = 2

2 = 2. Hence x = +7/√2 is a legitimate solution and not extraneous.

2. x = -7/√2

Substituting x = - 7/√2 in (x√2 - 5 = 2) we get,

(-7/√2)√2 - 5 = 2

-7 -5 = 2

-12 = 2. This is inconsistent as we know that -12 is not equal to 2.

Therefore x = -7/√2 is an extraneous solution.

## Given the equation √2x - 5 = 2, solve for x and identify if it is an extraneous solution.

**Summary:**

The equation √2x - 5 = 2 when solved for x has an extraneous solution x = -7/√2.

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