Question

If the sum and product of the roots of the quadratic equation \(kx^2+6x+4k=0\) are equal, then the value of k is

• A. \(-\frac{3}{2}\)
• B. \(\frac{3}{2}\)
• C. \(\frac{2}{3}\)
• D. \(-\frac{2}{3}\)

Answer 1

The Correct Option is A

We are given the quadratic equation: 

\( kx^2 + 6x + 4k = 0 \)

Let the roots be \( \alpha \) and \( \beta \).

From Vieta's formulas:

• Sum of the roots: \( \alpha + \beta = -\frac{\text{coefficient of } x}{\text{coefficient of } x^2} \)
• Product of the roots: \( \alpha \beta = \frac{\text{constant term}}{\text{coefficient of } x^2} \)

Substituting values:

\( \alpha + \beta = -\frac{6}{k} \)

\( \alpha \beta = \frac{4k}{k} = 4 \)

We are given that the sum and product of the roots are equal:

\( -\frac{6}{k} = 4 \)

Solving for \( k \):

\( k = -\frac{6}{4} = -\frac{3}{2} \)

Thus, the value of \( k \) is: \( -\frac{3}{2} \).

Answer 2

The given quadratic equation is \(kx^2+6x+4k=0\). 
According to Vieta's formulas, the sum of the roots \(\alpha + \beta\) is given by \(-\frac{b}{a}\) and the product of the roots \(\alpha \cdot \beta\) is given by \(\frac{c}{a}\), where \(a\), \(b\), and \(c\) correspond to the coefficients of the quadratic equation \(ax^2 + bx + c = 0\).

For our equation, \(a=k\), \(b=6\), and \(c=4k\). 

Thus, the sum of the roots is:

\[\alpha + \beta = -\frac{6}{k}\]

And the product of the roots is:

\[\alpha \cdot \beta = \frac{4k}{k} = 4\]

We are given that the sum and product of the roots are equal, i.e., \(\alpha + \beta = \alpha \cdot \beta\). 
Therefore,

\[-\frac{6}{k} = 4\]

Multiplying both sides by \(k\) gives us:

\[-6 = 4k\]

Solving for \(k\), we divide both sides by \(4\):

\[k = -\frac{6}{4} = -\frac{3}{2}\]

Hence, the value of \(k\) is \(-\frac{3}{2}\).