Verified answer

Step-by-step explanation:

1. Identify that α and β are roots of p(x).

2. Apply Vieta's formulas: α+β = k-6 and αβ = 2k+1.

3. Substitute α+β for αβ in the equation: α+β = αβ.

4. Plug in Vieta's formulas: k-6 = 2k+1.

5. Solve for k: k = -7.

Appropriate Question:

[tex]\sf \:If\: \alpha \: and \: \beta \: are \: zeroes \: of \: p(x) = x ^{2} - (k - 6)x + (2k + 1) \: \\ \sf \: and \: \alpha + \beta = \alpha \beta, \: then \: find \: the \: value \: of \: k. \qquad \: \qquad \: \: \: \: [/tex]

Answer:

[tex]\boxed{\bf \: k \: = \: - \: 7 \: } \\ [/tex]

Step-by-step explanation:

Given that,

[tex]\sf \: \alpha \: and \: \beta \: are \: zeroes \: of \: {x}^{2} - (k - 6)x + (2k + 1) \\ [/tex]

We know,

[tex]\boxed{{\sf Sum\ of\ the\ zeroes=\dfrac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}\\ [/tex]

[tex]\sf \:  \implies \: \alpha + \beta = - \dfrac{- (k -6)}{1} = k - 6 \\ [/tex]

Also,

[tex]\boxed{{\sf Product\ of\ the\ zeroes=\dfrac{Constant}{coefficient\ of\ x^{2}}}} \\ [/tex]

[tex]\sf \:  \implies \: \alpha \beta = \dfrac{2k + 1}{1} = 2k + 1 \\ [/tex]

Now, It is given that,

[tex]\sf \: \alpha + \beta = \alpha \beta \\ [/tex]

[tex]\sf \: k - 6 = 2k + 1 \\ [/tex]

[tex]\sf \: k - 2k = 6 + 1 \\ [/tex]

[tex]\sf \: - k =7 \\ [/tex]

[tex]\implies\sf \: \boxed{\bf \: k \: = \: - \: 7 \: } \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional Information:

[tex]\bf \:If\:\alpha, \beta \: are \: zeroes \: of \: {ax}^{2} + bx + c, \: then \\ [/tex]

[tex]\sf \: \alpha + \beta = - \dfrac{b}{a} \\ [/tex]

[tex]\sf \: \alpha \beta = \dfrac{c}{a} \\ [/tex]

[tex]\bf \:If\: \alpha, \beta, \gamma \: are \: zeroes \: of \: {ax}^{3} + {bx}^{2} + cx + d, \: then\\ [/tex]

[tex]\sf \: \alpha + \beta + \gamma = - \dfrac{b}{a} \\ [/tex]

[tex]\sf \: \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a}\\ [/tex]

[tex]\sf \: \alpha \beta \gamma = - \dfrac{d}{a} \\ [/tex]