Quadratic Equations Question - None

Question

Let n ∈ Z and △ABC be a right triangle with right angle at C. If sin A and sin B are the roots of the quadratic equation $(5n+8)x^2 - (7n-20)x + 120 = 0$, then find the value of n.

Accepted Answer

Step 1: In right triangle ABC with right angle at C, we have A + B = 90°, so B = 90° - A, which gives sin B = cos A.

Step 2: Let sin A = α and sin B = β. Then α and β are roots of $(5n+8)x^2 - (7n-20)x + 120 = 0$.

Step 3: By Vieta's formulas:

• Sum: $α + β = \frac{7n-20}{5n+8}$

• Product: $αβ = \frac{120}{5n+8}$

Step 4: Since sin B = cos A, we have β = cos A. Using sin²A + cos²A = 1:

$α^2 + β^2 = 1$

Step 5: Expand using the identity $α^2 + β^2 = (α+β)^2 - 2αβ$:

$(\frac{7n-20}{5n+8})^2 - 2·\frac{120}{5n+8} = 1$

Step 6: Let $5n+8 = d$. Then:

$\frac{(7n-20)^2}{d^2} - \frac{240}{d} = 1$

$(7n-20)^2 - 240d = d^2$

Step 7: Substitute $7n = \frac{5d-8}{5}·7 = 7n$ directly. With $d = 5n+8$:

$(7n-20)^2 - 240(5n+8) = (5n+8)^2$

$49n^2 - 280n + 400 - 1200n - 1920 = 25n^2 + 80n + 64$

$49n^2 - 1480n - 1520 = 25n^2 + 80n + 64$

$24n^2 - 1560n - 1584 = 0$

$n^2 - 65n - 66 = 0$

$(n-66)(n+1) = 0$

Step 8: Since $5n+8 > 0$ (coefficient of x²), we need $n > -\frac{8}{5}$. Thus $n = 66$.

∴ Answer: 66

Question:

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