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Question : 55 of 86

Marks: +1, -0

If

P + \sqrt{P} = \frac{6}{25}

P

is equal to

$\frac{1}{5}$
$\frac{2}{5}$
$\frac{1}{25}$
$\frac{2}{25}$

Solution:

Concept:

Substitute

\sqrt{P}=t

to convert into a quadratic equation.

Explanation:

Let

\sqrt{P}=t

, so

P=t^2

.

Equation:

t^2 + t = \frac{6}{25}

.

Multiply by 25:

25t^2 + 25t - 6 = 0

.

Discriminant:

25^2 - 4 \cdot 25 \cdot (-6) = 625 + 600 = 1225 = 35^2

.

Roots:

t = \frac{-25 \pm 35}{50}

.

t = \frac{10}{50} = \frac{1}{5}

or

t = \frac{-60}{50} = -\frac{6}{5}

.

Since

t = \sqrt{P} \ge 0

, take

t = \frac{1}{5}

.

Thus

P = t^2 = \left(\frac{1}{5}\right)^2 = \frac{1}{25}

.

Answer:

P = \frac{1}{25}

, which corresponds to option C.

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