Question

Let \(\alpha\) and \(\beta\) be the roots of \[ x^2+bx+c=0. \] If \[ \alpha^2+\beta^2=14 \] and \[ \alpha\beta=3, \] then the value of \(b^2\) is:

• A. \(8\)
• B. \(16\)
• C. \(20\)
• D. \(28\)

Hint:

Whenever a question involves roots of a quadratic equation, immediately write \[ \alpha+\beta=-\frac{\text{coefficient of }x}{\text{coefficient of }x^2}, \qquad \alpha\beta=\frac{\text{constant term}}{\text{coefficient of }x^2}. \] Most root-based problems become straightforward after applying these relations.

Answer 1

The Correct Option is C

Step 1: Recall the root relations.
For $x^2+bx+c=0$ with roots $\alpha$ and $\beta$, the sum of roots is $\alpha+\beta=-b$ and the product is $\alpha\beta=c$. These come straight from comparing coefficients.

Step 2: List what is given.
We know $\alpha^2+\beta^2=14$ and $\alpha\beta=3$. We want $b^2$.

Step 3: Use a square identity.
There is a handy identity: $(\alpha+\beta)^2=\alpha^2+\beta^2+2\alpha\beta$. It links the sum of squares to the sum and product of roots.

Step 4: Substitute the numbers.
Plug in: $(\alpha+\beta)^2=14+2(3)=14+6=20$.

Step 5: Connect to $b$.
Since $\alpha+\beta=-b$, squaring gives $(\alpha+\beta)^2=b^2$. So $b^2=20$.

Step 6: State the value.
Therefore $b^2=20$.

Step 7: Quick sanity check.
A positive value for $b^2$ makes sense because any real number squared is non-negative, and $20$ matches one of the options.
\[ \boxed{20} \]