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Use the discriminant to determine whether the quadratic equation has two unequal real solutions, a repeated real solution, or no real solution, without solving the equation. \( 2 x^{2}+7 x-9=0 \) Which of the following correctly describes the solutions to the given equation? A. A repeated real solution B. No real solution C. Two unequal real solutions

Ask by Schultz Allan. in the United States
Jan 24,2025

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The quadratic equation \( 2x^{2} + 7x - 9 = 0 \) has two unequal real solutions.

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Bonus Knowledge

To analyze the solutions of the quadratic equation \( 2 x^{2}+7 x-9=0 \), we first calculate the discriminant, which is given by the formula \( D = b^2 - 4ac \). Here, \( a = 2 \), \( b = 7 \), and \( c = -9 \). Plugging in those values, we get \( D = 7^2 - 4 \cdot 2 \cdot (-9) = 49 + 72 = 121 \). Since the discriminant is positive (\( D > 0 \)), this indicates that there are two unequal real solutions. Now, let's consider how this concept applies in real life. Quadratic equations pop up in various fields, from physics (think projectile motions!) to finance (calculating break-even points). By using the discriminant, you can quickly assess whether you need to find just one point of intersection or if you’ll be needing some calculations for multiple scenarios, making it a handy tool in both problem-solving and real-world applications!

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PHIẾU: PHÉP CỘNG, TRỪ PHÂN SỐ HỌ VÀ TÊN:.............................LỚP 8A4 Bài 1. Thực hiện phép tính: a) \( \frac{x-5}{5}+\frac{1-x}{5} \) b) \( \frac{x-y}{8}+\frac{2 y}{8} \) c) \( \frac{x^{2}+1}{x-2}-\frac{1-2 x}{x-2} \) d) \( \frac{4 x+1}{3}-\frac{x-2}{3} \) e) \( \frac{4 x-1}{3 x^{2} y}-\frac{7 x-1}{3 x^{2} y} \) f) \( \frac{3 x+2 y}{x-y}-\frac{2 x+3 y}{x-y} \) d) \( \frac{5 x y^{2}-x^{2} y}{3 x y}-\frac{4 x y^{2}+x^{2} y}{3 x y} \) e) \( \frac{x+1}{a-b}+\frac{x-1}{a-b}-\frac{x+3}{a-b} \) f) \( \frac{5 x y-4 y}{2 x^{2} y^{3}}+\frac{3 x y+4 y}{2 x^{2} y^{3}} \) h) \( \frac{x^{2}+4}{x-2}+\frac{4 x}{2-x} \) i) \( \frac{2 x^{2}-x y}{x-y}+\frac{x y+y^{2}}{y-x}-\frac{2 y^{2}-x^{2}}{x-y} \) Bài 2: Thực hiện phép tính: a) \( \frac{2 x+4}{10}+\frac{2-x}{15} \) b) \( \frac{x^{2}}{x^{2}+3 x}+\frac{3}{x+3}+\frac{3}{x} \) c) \( \frac{2}{x+y}-\frac{1}{y-x}+\frac{-3 x}{x^{2}-y^{2}} \) d) \( \frac{4}{x+2}+\frac{2}{x-2}+\frac{5 x-6}{4-x^{2}} \); e) \( \frac{1-3 x}{2 x}+\frac{3 x-2}{2 x-1}+\frac{3 x-2}{2 x-4 x^{2}} \); f) \( \frac{x^{2}+2}{x^{3}-1}+\frac{2}{x^{2}+x+1}+\frac{1}{1-x} \) Bài 3. Làm tính trừ các phân thức sau: a) \( \frac{4 x+1}{3}-\frac{x-2}{3} \) b) \( \frac{4 x-1}{3 x^{2} y}-\frac{7 x-1}{3 x^{2} y} \) c) \( \frac{3 x+2 y}{x-y}-\frac{2 x+3 y}{x-y} \) Bài 4. Làm các phép tính a) \( \frac{x y-1}{2 x-y}-\frac{1-2 x^{2}}{y-2 x} \) b) \( \frac{3 x y^{2}+x^{2} y}{x^{2} y-x y^{2}}-\frac{3 x^{2} y+x y^{2}}{x y(x-y)} \) c) \( \frac{x+9}{x^{2}-9}-\frac{3}{x^{2}+3 x} \) Bài 5. Thực hiện phép tính a) \( \frac{5 x^{2}}{6 x-6 y}-\frac{2 x^{2}}{3 y-3 x} \) b) \( \frac{y}{x y-5 x^{2}}-\frac{25 x-15 y}{25 x^{2}-y^{2}} \) c) \( \frac{1}{2 x-3}-\frac{2}{2 x+3}-\frac{6}{4 x^{2}-9} \) Bài 6. Rút gọn rồi tính giá trị của biểu thức a) \( \frac{x+1}{x-1}-\frac{4 x}{x^{2}-1} \) với \( x=-2 \) b) \( \frac{1}{x y-x^{2}}-\frac{1}{y^{2}-x y} \) với \( x=-5 ; y=\frac{-1}{5} \) Bài 7. Chứng minh biểu thức sau không phụ thuộc vào x : \[ A=\frac{11 x}{2 x-3}-\frac{x-18}{3-2 x} \quad B=\frac{1}{x+1}-\frac{2 x}{x-1}+\frac{x+3}{x^{2}-1} \] Bài 8. Chứng minh biểu thức sau không phụ thuộc vào \( \mathrm{x}, \mathrm{y} \) : \[ D=\frac{2 x+1}{x-2 y}-\frac{5 y+2}{x-2 y}-\frac{y+1}{2 y-x} \quad K=\frac{x+y}{x}-\frac{x}{x-y}+\frac{y^{2}}{x^{2}-x y} \]
Algebra Vietnam Jan 24, 2025
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