Question

If \( 2x^2 + 3x - 2 = 0 \) and \( 3x^2 + \alpha x - 2 = 0 \) have one common root, then the sum of all possible values of \( \alpha \) is:

• A. \( -3.5 \)
• B. \( 7.5 \)
• C. \( -7.5 \)
• D. \( -1.5 \)

Hint:

When two quadratic equations share a common root, subtract the equations to eliminate terms and solve for the unknown parameter. This can help find the possible values of the parameter.

Answer 1

The Correct Option is B

We are given two quadratic equations: \[ 2x^2 + 3x - 2 = 0 \quad \text{(Equation 1)}, \] \[ 3x^2 + \alpha x - 2 = 0 \quad \text{(Equation 2)}. \] We are told that these two equations have one common root, say \( r \). 
Step 1: 
Let the common root of both equations be \( r \). Substituting \( r \) into both equations: From Equation 1: \[ 2r^2 + 3r - 2 = 0. \tag{1} \] From Equation 2: \[ 3r^2 + \alpha r - 2 = 0. \tag{2} \] 
Step 2: 
Now, subtract Equation (1) from Equation (2) to eliminate the constant terms: \[ (3r^2 + \alpha r - 2) - (2r^2 + 3r - 2) = 0. \] Simplifying: \[ r^2 + (\alpha - 3)r = 0. \] Factoring: \[ r(r + \alpha - 3) = 0. \] 
Step 3: 
For this equation to hold, either \( r = 0 \) or \( r = 3 - \alpha \). - If \( r = 0 \), substitute \( r = 0 \) into Equation (1): \[ 2(0)^2 + 3(0) - 2 = -2 \neq 0. \] Therefore, \( r = 0 \) is not a valid root. - Thus, we must have \( r = 3 - \alpha \). 
Step 4: 
Substitute \( r = 3 - \alpha \) into Equation (1): \[ 2(3 - \alpha)^2 + 3(3 - \alpha) - 2 = 0. \] Expanding: \[ 2(9 - 6\alpha + \alpha^2) + 9 - 3\alpha - 2 = 0, \] \[ 18 - 12\alpha + 2\alpha^2 + 9 - 3\alpha - 2 = 0, \] \[ 2\alpha^2 - 15\alpha + 25 = 0. \] Now, solve this quadratic equation for \( \alpha \) using the quadratic formula: \[ \alpha = \frac{-(-15) \pm \sqrt{(-15)^2 - 4(2)(25)}}{2(2)} = \frac{15 \pm \sqrt{225 - 200}}{4} = \frac{15 \pm \sqrt{25}}{4} = \frac{15 \pm 5}{4}. \] Thus, the two possible values for \( \alpha \) are: \[ \alpha = \frac{15 + 5}{4} = \frac{20}{4} = 5, \quad \alpha = \frac{15 - 5}{4} = \frac{10}{4} = 2.5. \] 
Step 5: 
The sum of all possible values of \( \alpha \) is: \[ 5 + 2.5 = 7.5. \] Thus, the correct answer is: \[ \boxed{7.5}. \]

Answer 2

Given the two quadratic equations:
\[ 2x^2 + 3x - 2 = 0 \] and \[ 3x^2 + \alpha x - 2 = 0 \] which share one common root, find the sum of all possible values of \( \alpha \).

Step 1: Let the common root be \( r \). Then:
\[ 2r^2 + 3r - 2 = 0 \quad \text{and} \quad 3r^2 + \alpha r - 2 = 0 \]

Step 2: From the first equation:
\[ 2r^2 + 3r - 2 = 0 \] Multiply this equation by 3:
\[ 6r^2 + 9r - 6 = 0 \]

Step 3: From the second equation:
\[ 3r^2 + \alpha r - 2 = 0 \] Multiply this equation by 2:
\[ 6r^2 + 2 \alpha r - 4 = 0 \]

Step 4: Subtract the two equations:
\[ (6r^2 + 9r - 6) - (6r^2 + 2 \alpha r - 4) = 0 \] \[ 9r - 6 - 2 \alpha r + 4 = 0 \] \[ (9r - 2 \alpha r) - 2 = 0 \] \[ r (9 - 2 \alpha) = 2 \] \[ r = \frac{2}{9 - 2 \alpha} \]

Step 5: Substitute \( r = \frac{2}{9 - 2 \alpha} \) back into the first equation:
\[ 2 \left(\frac{2}{9 - 2 \alpha}\right)^2 + 3 \left(\frac{2}{9 - 2 \alpha}\right) - 2 = 0 \]

Step 6: Multiply both sides by \( (9 - 2 \alpha)^2 \) to clear denominators:
\[ 2 \times 4 + 3 \times 2 (9 - 2 \alpha) - 2 (9 - 2 \alpha)^2 = 0 \] \[ 8 + 6(9 - 2 \alpha) - 2 (9 - 2 \alpha)^2 = 0 \]

Step 7: Expand:
\[ 8 + 54 - 12 \alpha - 2 (81 - 36 \alpha + 4 \alpha^2) = 0 \] \[ 62 - 12 \alpha - 162 + 72 \alpha - 8 \alpha^2 = 0 \] \[ (62 - 162) + (-12 \alpha + 72 \alpha) - 8 \alpha^2 = 0 \] \[ -100 + 60 \alpha - 8 \alpha^2 = 0 \]

Step 8: Rearrange:
\[ -8 \alpha^2 + 60 \alpha - 100 = 0 \] Divide by -4:
\[ 2 \alpha^2 - 15 \alpha + 25 = 0 \]

Step 9: Solve quadratic for \( \alpha \):
\[ \alpha = \frac{15 \pm \sqrt{(-15)^2 - 4 \times 2 \times 25}}{2 \times 2} = \frac{15 \pm \sqrt{225 - 200}}{4} = \frac{15 \pm \sqrt{25}}{4} \] \[ \alpha = \frac{15 \pm 5}{4} \]

Step 10: Find the two values:
\[ \alpha_1 = \frac{15 + 5}{4} = \frac{20}{4} = 5 \] \[ \alpha_2 = \frac{15 - 5}{4} = \frac{10}{4} = 2.5 \]

Step 11: Sum of all possible values:
\[ 5 + 2.5 = 7.5 \]

Therefore, the sum of all possible values of \( \alpha \) is:
\[ \boxed{7.5} \]