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

Concept: A quadratic equation \[ ax^2+bx+c=0 \] has real roots if and only if its discriminant is non-negative. The discriminant is: \[ D=b^2-4ac \] For the equation \[ x^2+bx+c=0, \] we have: \[ a=1 \] Hence, \[ D=b^2-4c \] The equation will have real roots whenever: \[ b^2-4c \geq 0 \] That is, \[ b^2 \geq 4c \] We now count all possible ordered pairs \( (b,c) \) satisfying this condition.

Step 1: Find total number of possible selections.
Since both \( b \) and \( c \) are selected from: \[ \{1,2,3,\ldots,10\} \] with replacement, Number of choices for \( b = 10 \) Number of choices for \( c = 10 \) Therefore, \[ \text{Total outcomes}=10\times10=100 \]

Step 2: Apply the condition \( b^2 \geq 4c \).
We evaluate possible values of \( c \) for each value of \( b \). \[ \begin{array}{|c|c|c|} \hline b & b^2 & c \text{ satisfying } c\le \frac{b^2}{4} \\ \hline 1 & 1 & 0 \\ 2 & 4 & 1 \\ 3 & 9 & 2 \\ 4 & 16 & 4 \\ 5 & 25 & 6 \\ 6 & 36 & 9 \\ 7 & 49 & 10 \\ 8 & 64 & 10 \\ 9 & 81 & 10 \\ 10 & 100 & 10 \\ \hline \end{array} \] Now count total favorable cases: \[ 0+1+2+4+6+9+10+10+10+10 \] \[ =62 \]

Step 3: Compute the probability.
\[ P(\text{real roots}) = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} \] \[ = \frac{62}{100} \] \[ =0.62 \] Thus, \[ \boxed{0.62} \] Hence the correct option is: \[ \boxed{(D)} \]