Question

Let the value of \(p\), such that the sum of the squares of the roots of the quadratic equation \[ x^2+(7-p)x+4=p \] has the least value, be \(\alpha\), and the corresponding roots be \(\beta\) and \(\gamma\). Then \(\alpha^3+\beta^3+\gamma^3\) equals \underline{\hspace{2cm}.}

Hint:

Whenever a question asks for the minimum or maximum value involving roots, express everything in terms of coefficients and reduce it to a quadratic form.

Answer 1

Correct Answer: 209

Concept:
If the roots of a quadratic equation are \(\beta\) and \(\gamma\), then \[ \beta+\gamma = -\frac{b}{a}, \quad \beta\gamma=\frac{c}{a}. \]
The sum of squares of roots is \[ \beta^2+\gamma^2=(\beta+\gamma)^2-2\beta\gamma. \]
To find the least value of an expression involving a parameter, convert it into a quadratic in that parameter and minimize it.
Step 1: Write the quadratic equation in standard form \[ x^2+(7-p)x+(4-p)=0 \]
Step 2: Find sum and product of roots \[ \beta+\gamma=p-7,\qquad \beta\gamma=4-p \]
Step 3: Find the sum of squares of the roots \[ \beta^2+\gamma^2=(p-7)^2-2(4-p) \] \[ = p^2-14p+49-8+2p \] \[ = p^2-12p+41 \]
Step 4: Minimize the sum of squares The expression \[ p^2-12p+41 \] is minimum when \[ p=\frac{12}{2}=6 \] Hence, \[ \alpha=6 \]
Step 5: Find \(\beta+\gamma\) and \(\beta\gamma\) for \(p=6\) \[ \beta+\gamma=6-7=-1,\qquad \beta\gamma=4-6=-2 \]
Step 6: Find \(\beta^3+\gamma^3\) \[ \beta^3+\gamma^3=(\beta+\gamma)^3-3\beta\gamma(\beta+\gamma) \] \[ =(-1)^3-3(-2)(-1)=-1-6=-7 \]
Step 7: Find the required value \[ \alpha^3+\beta^3+\gamma^3=6^3-7=216-7=209 \]
Final Answer: \(\boxed{209}\)