Question

If \( b \) and \( c \) are numbers chosen at random from the set \( \{1,2,3,\ldots,10\} \) with replacement, then the probability that the quadratic equation \[ x^2 + bx + c = 0 \] has real roots is:

• A. \(0.52\)
• B. \(0.54\)
• C. \(0.58\)
• D. \(0.62\)

Hint:

For quadratic equations: \[ ax^2+bx+c=0 \] always remember: \[ \text{Real roots} \iff b^2-4ac \geq 0 \] In probability questions involving coefficients, systematically count all valid cases satisfying the discriminant condition.

Answer 1

The Correct Option is D

Step 1: Condition for real roots.
For $x^2+bx+c=0$ to have real roots, the discriminant must be non-negative: $b^2-4c\ge0$, that is $b^2\ge4c$.
Step 2: Count total outcomes.
Both $b$ and $c$ come from $\{1,\dots,10\}$ with replacement, so there are $10\times10=100$ ordered pairs.
Step 3: For each $b$, count valid $c$.
Valid $c$ are those with $c\le\dfrac{b^2}{4}$. For $b=1$: $0$; $b=2$: $1$; $b=3$: $2$; $b=4$: $4$; $b=5$: $6$.
Step 4: Continue the count.
$b=6$: $9$; $b=7$: $10$; $b=8$: $10$; $b=9$: $10$; $b=10$: $10$ (capped at $10$ since $c\le10$).
Step 5: Add the favourable cases.
$0+1+2+4+6+9+10+10+10+10=62$.
Step 6: Find the probability.
$P=\dfrac{62}{100}=0.62$. \[ \boxed{0.62} \]