Algebra

Polynomials

Medium

If $8x^2 - 2kx + k = 0$ is a quadratic equation in $x$ , such that one of its roots is $p$ times the other, and $p, k$ are positive real numbers, then $k$ equals

$(p + \frac{1}{p})$

$2(p + \frac{1}{p})$

$2(\sqrt{p} + \frac{1}{\sqrt{p}})^2$

$(\sqrt{p} + \frac{1}{\sqrt{p}})^2$

✅ Correct Option: 3

Let the two roots of the quadratic equation $8x^2 - 2kx + k = 0$ be $\alpha$ and $p\alpha$ , where one root is $p$ times the other.

For a quadratic equation $ax^2 + bx + c = 0$ :

Sum of roots $= -\frac{b}{a}$

Product of roots $= \frac{c}{a}$

In our equation, $a = 8$ , $b = -2k$ , $c = k$

Applying sum of roots:

$\alpha + p\alpha = -\frac{(-2k)}{8}$

$\alpha(1 + p) = \frac{k}{4}$ ... (1)

Applying product of roots:

$\alpha \times p\alpha = \frac{k}{8}$

$p\alpha^2 = \frac{k}{8}$ ... (2)

From equation (1):

$\alpha = \frac{k}{4(1 + p)}$

Substituting this into equation (2):

$p \times \left(\frac{k}{4(1 + p)}\right)^2 = \frac{k}{8}$

$p \times \frac{k^2}{16(1 + p)^2} = \frac{k}{8}$

Since $k > 0$ , dividing both sides by $k$ :

$\frac{pk}{16(1 + p)^2} = \frac{1}{8}$

Cross-multiplying:

$8pk = 16(1 + p)^2$

$pk = 2(1 + p)^2$

$k = \frac{2(1 + p)^2}{p}$

Expanding:

$k = \frac{2(1 + 2p + p^2)}{p} = \frac{2}{p} + 4 + 2p$

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If $8x^2 - 2kx + k = 0$ is a quadratic equation in $x$ , such that one of its roots is $p$ times the other, and $p, k$ are positive real numbers, then $k$ equals