SOLUTION: Find all nonzero constants $a$ such that
ax^2 + 7x + 2 = 5x^2 + 23x - 12
has only one distinct root.
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Question 1209641: Find all nonzero constants $a$ such that
ax^2 + 7x + 2 = 5x^2 + 23x - 12
has only one distinct root.
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Here's how to find the values of *a* for which the given quadratic equation has only one distinct root:
1. **Rewrite the equation in standard form:**
Combine like terms to get the equation in the form Ax² + Bx + C = 0:
(a - 5)x² + (7 - 23)x + (2 + 12) = 0
(a - 5)x² - 16x + 14 = 0
2. **Apply the discriminant condition:**
A quadratic equation has only one distinct root (a double root) when its discriminant (B² - 4AC) is equal to zero. In our equation:
* A = (a - 5)
* B = -16
* C = 14
So, we set the discriminant equal to zero:
(-16)² - 4 * (a - 5) * 14 = 0
256 - 56(a - 5) = 0
256 - 56a + 280 = 0
536 - 56a = 0
3. **Solve for *a*:**
56a = 536
a = 536 / 56
a = 67/7
Therefore, the only nonzero constant *a* for which the given equation has only one distinct root is a = 67/7.
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