Consider the following quadratic equations:

Equation 1: $ax^2 + bx + c = 0$

Equation 2: $dy^2 + ey + c = 0$

Given the conditions:

1. $c$ is a single-digit prime number greater than 2.
2. $d$ and $b$ are two-digit prime numbers less than 20.
3. $d$ is greater than 11.
4. $b$ is greater than $d$.
5. The smallest roots of both equations are equal.
6. Neither equation has irrational roots.
7. $e = b + 1$.
8. $3c$ is greater than $b$.

Find the value of $d imes e$.

Question

Consider the following quadratic equations:

Equation 1: $ax^2 + bx + c = 0$

Equation 2: $dy^2 + ey + c = 0$

Given the conditions:

1. $c$ is a single-digit prime number greater than 2.
2. $d$ and $b$ are two-digit prime numbers less than 20.
3. $d$ is greater than 11.
4. $b$ is greater than $d$.
5. The smallest roots of both equations are equal.
6. Neither equation has irrational roots.
7. $e = b + 1$.
8. $3c$ is greater than $b$.

Find the value of $d 	imes e$.

Accepted Answer

**Solution:**

1. From the problem, $c$ is a single-digit prime number greater than 2. The possible values for $c$ are therefore $3, 5,$ or $7$.

2. $d$ and $b$ are two-digit prime numbers less than 20, with $d > 11$ and $b > d$. The two-digit primes less than 20 are $11, 13, 17, 19$. Since $d > 11$, possible values for $d$ are $13, 17,$ or $19$. Since $b > d$, $b$ must be a prime greater than $d$ from the same set.

3. Condition $3c > b$ restricts $b$ further.

4. Since $e = b + 1$, $e$ is one more than $b$.

5. Both equations have the same smallest root, and no roots are irrational. This means the discriminants of both quadratics are perfect squares.

6. Let's check possible values:

- Try $c = 5$ (since $3c = 15$).

- For $d$, possible values: $13, 17, 19$.

- For $b$, possible values greater than $d$ and less than $3c=15$:

- If $d=13$, $b$ can be $17$ or $19$, but $3c=15$ must be greater than $b$, so $b$ cannot be $17$ or $19$.

- If $d=13$, $b$ cannot be $17$ or $19$ because $3c=15$ is not greater than $17$ or $19$.

- Try $c=7$, then $3c=21$.

- For $d=13$, $b$ can be $17$ or $19$ (both less than 21).

- Check for $d=13$, $b=19$:

- $e = b + 1 = 20$.

7. Now, check the roots condition:

- Equation 1: $ax^2 + 19x + 7 = 0$

- Equation 2: $13y^2 + 20y + 7 = 0$

8. Since the smallest roots are equal, let the common smallest root be $r$.

9. For quadratic $ax^2 + bx + c = 0$, roots are given by:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

Similarly for the second equation.

10. Since roots are rational, discriminants must be perfect squares.

- For equation 2:

$$\Delta_2 = e^2 - 4dc = 20^2 - 4 	imes 13 	imes 7 = 400 - 364 = 36$$

which is a perfect square.

- For equation 1:

$$\Delta_1 = b^2 - 4ac = 19^2 - 4a 	imes 7 = 361 - 28a$$

This must be a perfect square.

11. Let $361 - 28a = k^2$, where $k$ is an integer.

Try values of $a$ such that $k^2$ is a perfect square less than 361:

- For $a=9$:

$$361 - 28 	imes 9 = 361 - 252 = 109$$ (not a perfect square)

- For $a=10$:

$$361 - 280 = 81$$ (which is $9^2$, a perfect square)

So, $a=10$ works.

12. Now, roots of equation 1:

$$x = \frac{-19 \pm 9}{2 	imes 10}$$

Roots are:

- $$\frac{-19 - 9}{20} = \frac{-28}{20} = -1.4$$

- $$\frac{-19 + 9}{20} = \frac{-10}{20} = -0.5$$

13. Roots of equation 2:

$$y = \frac{-20 \pm 6}{2 	imes 13}$$

Roots are:

- $$\frac{-20 - 6}{26} = \frac{-26}{26} = -1$$

- $$\frac{-20 + 6}{26} = \frac{-14}{26} = -\frac{7}{13} \approx -0.538$$

14. The smallest root of both equations must be the same. Compare the roots:

- Equation 1 smallest root: $-1.4$

- Equation 2 smallest root: $-1$

They are not equal.

Try $b=17$:

- $e = 18$

- Equation 2 discriminant:

$$18^2 - 4 	imes 13 	imes 7 = 324 - 364 = -40$$

Negative, discard.

Try $d=17$, $b=19$:

- $e = 20$

- Equation 2 discriminant:

$$20^2 - 4 	imes 17 	imes 7 = 400 - 476 = -76$$

Negative, discard.

Try $d=19$, $b= 17$ (not possible since $b > d$ is given).

Try $c=3$, $3c=9$:

- $b < 9$, $b$ two-digit prime less than 20 and greater than $d$ (which is greater than 11), no such $b$.

Hence, the only possible pair is $d=13$, $b=19$, $e=20$, $c=7$.

Calculate $d 	imes e = 13 		imes 20 = 260$.

**Answer:** $260$

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