Multiple choice

if $ { \alpha }{ 1 },{ \alpha }{ 2 }$ are the roots of the equation $ { x^{ 2 } }+px+q=0 $ and $ { \beta }{ 1 },{ \beta }{ 2 }$ are those of the equation $ { x^{ 2 } }+rx+s=0$ and the system of the equation $ { \alpha }{ 1 }y+{ \alpha }{ 2 }z=0 $ and $ { \beta }{ 1 }y+{ \beta }{ 2 }z=0 $ has non trivial solution, then

  1. $ \dfrac { q^{ 2 } }{ r^{ 2 } } =\dfrac { p}{ s } $
  2. $ \dfrac { p^{ 2 } }{ q^{ 2 } } =\dfrac { r }{ s } $
  3. $ \dfrac { p^{ 2 } }{ s^{ 2 } } =\dfrac { q }{ r } $
  4. $ \dfrac { p^{ 2 } }{ r^{ 2 } } =\dfrac { q }{ s } $
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D Correct answer
Explanation

A system of linear equations has a non-trivial solution if the determinant of the coefficient matrix is zero: alpha1*beta2 - alpha2*beta1 = 0. This implies alpha1/alpha2 = beta1/beta2. Using the properties of roots for the given quadratic equations, this leads to the relation p^2/r^2 = q/s.

AI explanation

For the homogeneous system of linear equations alpha_1 y + alpha_2 z = 0 and beta_1 y + beta_2 z = 0 to have a non-trivial solution, the determinant of its coefficient matrix must be zero. This gives the condition alpha_1 * beta_2 - alpha_2 * beta_1 = 0, which means the ratio of the roots alpha_1 / alpha_2 equals the ratio beta_1 / beta_2. Using the sum and product of roots formulas, the ratio of the first root to the second root for x^2 + px + q = 0 is (-p + sqrt(p^2 - 4q)) / (-p - sqrt(p^2 - 4q)), and for the second equation it is (-r + sqrt(r^2 - 4s)) / (-r - sqrt(r^2 - 4s)). Equating these ratios and cross-multiplying leads to the algebraic identity p^2 / r^2 = q / s.

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