If x = 1, y = 2 is a solution of the equation a²x + ay = 3, then find the values of a :

Q: If x = 1, y = 2 is a solution of the equation a²x + ay = 3, then find the values of a :

To determine the values of $ a $ for which $ x = 1, y = 2 $ is a solution of the equation $ a^2x + ay = 3 $, we substitute $ x = 1 $ and $ y = 2 $ into the equation: \[ a^2(1) + a(2) = 3 \] This simplifies to: \[ a^2 + 2a = 3 \] Rearranging the equation gives: \[ a^2 + 2a - 3 = 0 \] This is a quadratic equation in the standard form $ ax^2 + bx + c = 0 $, where $ a = 1 $, $ b = 2 $, and $ c = -3 $. To find the values of $ a $, we solve this quadratic equation using the quadratic formula: \[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Substituting the values of $ b $, $ a $, and $ c $: \[ a = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1} \] \[ a = \frac{-2 \pm \sqrt{4 + 12}}{2} \] \[ a = \frac{-2 \pm \sqrt{16}}{2} \] \[ a = \frac{-2 \pm 4}{2} \] This gives two solutions: 1. $ a = \frac{-2 + 4}{2} = \frac{2}{2} = 1 $ 2. $ a = \frac{-2 - 4}{2} = \frac{-6}{2} = -3 $ Thus, the values of $ a $ are $ 1 $ and $ -3 $, which corresponds to Option 1: $ a = 1, -3 $. Now, let's explain why the other options are incorrect: - **Option 2: $ a = 1, 2 $**: This option is incorrect because when solving the quadratic equation, the value $ a = 2 $ does not satisfy the equation $ a^2 + 2a - 3 = 0 $. - **Option 3: $ a = -3, 2 $**: This option is incorrect because, although $ a = -3 $ is a correct solution, $ a = 2 $ does not satisfy the equation. - **Option 4: $ a = -1, -2 $**: This option is incorrect because neither $ a = -1 $ nor $ a = -2 $ satisfies the equation $ a^2 + 2a - 3 = 0 $. Therefore, the correct answer is Option 1: $ a = 1, -3 $.

Question

Accepted Answer

To determine the values of $ a $ for which $ x = 1, y = 2 $ is a solution of the equation $ a^2x + ay = 3 $, we substitute $ x = 1 $ and $ y = 2 $ into the equation:

$$ a^2(1) + a(2) = 3 $$

This simplifies to:

$$ a^2 + 2a = 3 $$

Rearranging the equation gives:

$$ a^2 + 2a - 3 = 0 $$

This is a quadratic equation in the standard form $ ax^2 + bx + c = 0 $, where $ a = 1 $, $ b = 2 $, and $ c = -3 $. To find the values of $ a $, we solve this quadratic equation using the quadratic formula:

$$ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

Substituting the values of $ b $, $ a $, and $ c $:

$$ a = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1} $$

$$ a = \frac{-2 \pm \sqrt{4 + 12}}{2} $$

$$ a = \frac{-2 \pm \sqrt{16}}{2} $$

$$ a = \frac{-2 \pm 4}{2} $$

This gives two solutions:

1. $ a = \frac{-2 + 4}{2} = \frac{2}{2} = 1 $
2. $ a = \frac{-2 - 4}{2} = \frac{-6}{2} = -3 $

Thus, the values of $ a $ are $ 1 $ and $ -3 $, which corresponds to Option 1: $ a = 1, -3 $.

Now, let's explain why the other options are incorrect:

- **Option 2: $ a = 1, 2 $**: This option is incorrect because when solving the quadratic equation, the value $ a = 2 $ does not satisfy the equation $ a^2 + 2a - 3 = 0 $.

- **Option 3: $ a = -3, 2 $**: This option is incorrect because, although $ a = -3 $ is a correct solution, $ a = 2 $ does not satisfy the equation.

- **Option 4: $ a = -1, -2 $**: This option is incorrect because neither $ a = -1 $ nor $ a = -2 $ satisfies the equation $ a^2 + 2a - 3 = 0 $.

Therefore, the correct answer is Option 1: $ a = 1, -3 $.