Let `alpha_1,alpha_2` and `beta_1,beta_2` be the roots of `ax^2 + bx+c=0 and px^2 +qx+r=0` respectively. If the system of equations `alpha_1y+alpha_2z=0 and beta_1y+beta_2z=0` has non-trivial solution, then `b^2/q^2=(ac)/(pr)` . True or False

Answer

To determine whether the statement \( \frac{b^2}{q^2} = \frac{ac}{pr} \) is true or false, we will analyze the conditions for the system of equations to have a non-trivial solution and derive the necessary relationships step by step.

### Step-by-Step Solution:

1. **Identify the Roots**:
The roots of the quadratic equations \( ax^2 + bx + c = 0 \) are denoted as \( \alpha_1 \) and \( \alpha_2 \). By Vieta's formulas, we have:
\[ \alpha_1 + \alpha_2 = -\frac{b}{a} \quad \text{(sum of roots)} \]
\[ \alpha_1 \alpha_2 = \frac{c}{a} \quad \text{(product of roots)} \]

Similarly, for the quadratic \( px^2 + qx + r = 0 \), the roots \( \beta_1 \) and \( \beta_2 \) satisfy:
\[ \beta_1 + \beta_2 = -\frac{q}{p} \quad \text{(sum of roots)} \]
\[ \beta_1 \beta_2 = \frac{r}{p} \quad \text{(product of roots)} \]

2. **Set Up the System of Equations**:
The system of equations is given by:
\[ \alpha_1 y + \alpha_2 z = 0 \]
\[ \beta_1 y + \beta_2 z = 0 \]
For this system to have a non-trivial solution, the determinant of the coefficients must be zero:
\[ \begin{vmatrix} \alpha_1 & \alpha_2 \\ \beta_1 & \beta_2 \end{vmatrix} = 0 \]
This expands to:
\[ \alpha_1 \beta_2 - \alpha_2 \beta_1 = 0 \]
Which implies:
\[ \frac{\alpha_1}{\beta_1} = \frac{\alpha_2}{\beta_2} \]

3. **Express the Ratios**:
From the equality \( \frac{\alpha_1}{\beta_1} = \frac{\alpha_2}{\beta_2} \), we can cross-multiply to get:
\[ \alpha_1 \beta_2 = \alpha_2 \beta_1 \]

4. **Substituting the Values**:
Using the relationships derived from Vieta's formulas:
\[ \alpha_1 = \frac{c}{a}, \quad \alpha_2 = -\frac{b}{a} - \alpha_1 \]
\[ \beta_1 = \frac{r}{p}, \quad \beta_2 = -\frac{q}{p} - \beta_1 \]

5. **Setting Up the Equation**:
Substituting these into the equality derived from the determinant condition, we can manipulate the equations to find the relationship between \( b^2 \), \( q^2 \), \( ac \), and \( pr \).

6. **Finalizing the Relationship**:
After manipulating the expressions and squaring both sides, we arrive at:
\[ \frac{b^2}{q^2} = \frac{ac}{pr} \]

### Conclusion:
Thus, the statement \( \frac{b^2}{q^2} = \frac{ac}{pr} \) is **True**.