Given the two quadratic equations:

$$a^2 - 7a + d = 0$$

$$b^2 - 4b + (d - 9) = 0$$

where the roots of the first equation are $x$ and $y$, and the roots of the second equation are $x$ and $y - x$.

If we multiply the second equation by $3z$ and add $1$ to the smallest root of the newly formed equation, what is the resulting number?

Question

Given the two quadratic equations:

$$a^2 - 7a + d = 0$$

$$b^2 - 4b + (d - 9) = 0$$

where the roots of the first equation are $x$ and $y$, and the roots of the second equation are $x$ and $y - x$.

If we multiply the second equation by $3z$ and add $1$ to the smallest root of the newly formed equation, what is the resulting number?

Accepted Answer

**Solution:**

1. Let the roots of the first equation $a^2 - 7a + d = 0$ be $x$ and $y$.
   - Sum of roots: $x + y = 7$
   - Product of roots: $xy = d$

2. The roots of the second equation $b^2 - 4b + (d - 9) = 0$ are given as $x$ and $y - x$.
   - Sum of roots: $x + (y - x) = y = 4$ (from the equation's coefficient)
   - Product of roots: $x(y - x) = d - 9$

3. From step 1, $x + y = 7$, and from step 2, $y = 4$.
   - Therefore, $x + 4 = 7 ightarrow x = 3$

4. Using the product of roots from the first equation:
   - $xy = d ightarrow 3 	imes 4 = d = 12$

5. Using the product of roots from the second equation:
   - $x(y - x) = d - 9$
   - $3(4 - 3) = 12 - 9$
   - $3 	imes 1 = 3$
   - $3 = 3$ (consistent)

6. Now, the second equation is:
   $$b^2 - 4b + (12 - 9) = b^2 - 4b + 3 = 0$$

7. Multiply this equation by $3z$ (assuming $z=1$ as $z$ is not defined, or considering the multiplier as $3$):
   $$3 	imes (b^2 - 4b + 3) = 3b^2 - 12b + 9 = 0$$

8. The roots of this new equation are the same as the original second equation, but scaled. The roots remain the same.

9. The smallest root of the original second equation is $x = 3$ and $y - x = 4 - 3 = 1$, so smallest root is $1$.

10. Adding $1$ to the smallest root:
    $$1 + 1 = 2$$

**Answer:** $2$

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