9y² - 12y + 2 = 0
Understand the Problem
प्रश्न हे एक गणितीय समस्या आहे ज्यात आपण विशिष्ट समीकरणे सोडवण्याबद्दल विचारत आहात. यात एक त्रिकोणात्मक समीकरण दिले आहे ज्याचे निराकरण करणे आवश्यक आहे.
Answer
The roots are \( y_1 = \frac{2 + \sqrt{2}}{3} \) and \( y_2 = \frac{2 - \sqrt{2}}{3} \).
Answer for screen readers
The roots of the equation (9y^2 - 12y + 2 = 0) are: $$ y_1 = \frac{2 + \sqrt{2}}{3}, \quad y_2 = \frac{2 - \sqrt{2}}{3} $$
Steps to Solve
- Identify the quadratic equation
Given the quadratic equation is: $$9y^2 - 12y + 2 = 0$$
- Apply the quadratic formula
The quadratic formula to find the roots of the equation $ax^2 + bx + c = 0$ is: $$ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ Where (a = 9), (b = -12), and (c = 2).
- Calculate the discriminant
Calculate the discriminant using: $$ D = b^2 - 4ac $$ Substituting the values: $$ D = (-12)^2 - 4 \cdot 9 \cdot 2 = 144 - 72 = 72 $$
- ** Calculate the roots**
Now substitute (D) back into the quadratic formula: $$ y = \frac{-(-12) \pm \sqrt{72}}{2 \cdot 9} $$ Calculate further: $$ y = \frac{12 \pm 6\sqrt{2}}{18} $$ This can be simplified to: $$ y = \frac{2 \pm \sqrt{2}}{3} $$
- Write the final roots
The solutions are: $$ y_1 = \frac{2 + \sqrt{2}}{3}, \quad y_2 = \frac{2 - \sqrt{2}}{3} $$
The roots of the equation (9y^2 - 12y + 2 = 0) are: $$ y_1 = \frac{2 + \sqrt{2}}{3}, \quad y_2 = \frac{2 - \sqrt{2}}{3} $$
More Information
These roots represent the values of (y) that satisfy the given quadratic equation. The discriminant (D) is positive, indicating two distinct real roots.
Tips
- Not simplifying the square root term: Make sure to simplify $\sqrt{D}$ properly.
- Sign errors: Pay attention to the signs of (b) and whether it is being negated correctly.
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