Solving Quadratic Equations with Root Differences

The given quadratic equation is $\small 4x^2 - (5k + 1)x + 5k = 0$.

This equation is in the standard form $\small ax^2 + bx + c = 0$, where:

• $\small a = 4$
• $\small b = -(5k + 1)$
• $\small c = 5k$

Let the roots of the equation be $\small \alpha$ and $\small \beta$.

We are given that the roots differ by unity, which means the absolute difference between the roots is 1. Mathematically, this can be written as $\small |\alpha - \beta| = 1$.

Using Root Properties of Quadratic Equations

For a quadratic equation $\small ax^2 + bx + c = 0$, the sum and product of the roots are given by:

• Sum of roots: $\small \alpha + \beta = -\frac{b}{a}$
• Product of roots: $\small \alpha \beta = \frac{c}{a}$

Using the coefficients from our equation:

• Sum of roots: $\small \alpha + \beta = -\frac{-(5k + 1)}{4} = \frac{5k + 1}{4}$
• Product of roots: $\small \alpha \beta = \frac{5k}{4}$

Applying the Difference Condition

We know the identity relating the square of the difference of roots to their sum and product:

$\small (\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$

We are given $\small |\alpha - \beta| = 1$, so squaring both sides gives $\small (\alpha - \beta)^2 = 1^2 = 1$.

Substitute the expressions for the sum and product of roots in terms of k into the identity:

$\small 1 = \left(\frac{5k + 1}{4}\right)^2 - 4\left(\frac{5k}{4}\right)$

Simplify the equation:

$\small 1 = \frac{(5k + 1)^2}{16} - 5k$

To eliminate the fraction, multiply the entire equation by 16:

$\small 1 \times 16 = 16 \times \left(\frac{(5k + 1)^2}{16}\right) - 16 \times (5k)$

$\small 16 = (5k + 1)^2 - 80k$

Expand the term $\small (5k + 1)^2$:

$\small 16 = (25k^2 + 10k + 1) - 80k$

Combine like terms:

$\small 16 = 25k^2 - 70k + 1$

Rearrange the equation to form a standard quadratic equation in terms of k:

$\small 25k^2 - 70k + 1 - 16 = 0$

$\small 25k^2 - 70k - 15 = 0$

We can simplify this quadratic equation by dividing by 5:

$\small \frac{25k^2}{5} - \frac{70k}{5} - \frac{15}{5} = 0$

$\small 5k^2 - 14k - 3 = 0$

Solving for the Possible Values of k

We now have a quadratic equation in k: $\small 5k^2 - 14k - 3 = 0$. We can solve this by factoring.

We look for two numbers that multiply to $\small 5 \times (-3) = -15$ and add up to $\small -14$. These numbers are -15 and 1.

Rewrite the middle term using these numbers:

$\small 5k^2 - 15k + k - 3 = 0$

Group terms and factor:

$\small 5k(k - 3) + 1(k - 3) = 0$

Factor out the common term $\small (k - 3)$:

$\small (5k + 1)(k - 3) = 0$

For the product of two factors to be zero, at least one of the factors must be zero.

Case 1: $\small 5k + 1 = 0$

$\small 5k = -1$

$\small k = -\frac{1}{5}$

Case 2: $\small k - 3 = 0$

$\small k = 3$

So, the possible values of k are $\small -\frac{1}{5}$ and $\small 3$.

Comparing with Options

We need to find which of the given options is one of these possible values of k.

• Option 1: -3 (Not $\small -\frac{1}{5}$ or 3)
• Option 2: -1 (Not $\small -\frac{1}{5}$ or 3)
• Option 3: $\small -\frac{1}{5}$ (This is one of the possible values)
• Option 4: $\small -\frac{3}{5}$ (Not $\small -\frac{1}{5}$ or 3)

The possible value of k among the options is $\small -\frac{1}{5}$.

Revision Table: Quadratic Roots Properties

Additional Information: Quadratic Equations

A quadratic equation is a polynomial equation of the second degree. The general form is $\small ax^2 + bx + c = 0$, where $\small a, b, c$ are coefficients and $\small a \neq 0$.

The discriminant of a quadratic equation is $\small \Delta = b^2 - 4ac$. The discriminant tells us about the nature of the roots:

• If $\small \Delta > 0$, there are two distinct real roots.
• If $\small \Delta = 0$, there is exactly one real root (a repeated root).
• If $\small \Delta < 0$, there are two complex roots.

The roots of a quadratic equation can be found using the quadratic formula:

$\small x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In our problem, the condition that the roots differ by unity implies that the roots must be real, so the discriminant must be non-negative. Let's check this for the found values of k.

• For $\small k = -\frac{1}{5}$: $\small b = -(5(-\frac{1}{5}) + 1) = -(-1 + 1) = 0$, $\small c = 5(-\frac{1}{5}) = -1$. The equation is $\small 4x^2 - 1 = 0$. Roots are $\small x = \pm \frac{1}{2}$. Difference is $\small |\frac{1}{2} - (-\frac{1}{2})| = |1| = 1$. Discriminant $\small \Delta = 0^2 - 4(4)(-1) = 16 > 0$. (Correction: Discriminant for $4x^2 - 1 = 0$ is $b^2-4ac = 0^2 - 4(4)(-1) = 16 > 0$, which gives real roots. The roots are $\pm 1/2$, difference is 1. This value of k is valid).
• For $\small k = 3$: $\small b = -(5(3) + 1) = -(15 + 1) = -16$, $\small c = 5(3) = 15$. The equation is $\small 4x^2 - 16x + 15 = 0$. Discriminant $\small \Delta = (-16)^2 - 4(4)(15) = 256 - 240 = 16 > 0$. Roots are real. Using the quadratic formula: $\small x = \frac{-(-16) \pm \sqrt{16}}{2(4)} = \frac{16 \pm 4}{8}$. Roots are $\small x_1 = \frac{16 + 4}{8} = \frac{20}{8} = \frac{5}{2}$ and $\small x_2 = \frac{16 - 4}{8} = \frac{12}{8} = \frac{3}{2}$. The difference is $\small |\frac{5}{2} - \frac{3}{2}| = |\frac{2}{2}| = |1| = 1$. This value of k is also valid.

Both values of k lead to real roots that differ by unity.