Multiple choice

Let ${ \alpha }{ 1 },{ \alpha }{ 2 }$ and ${ \beta }{ 1 },{ \beta }{ 2 }$ be the roots of $a{ x }^{ 2 }+bx+c=0$ and $p{ x }^{ 2 }+qx+r=0$ respectively. If the system of equations ${ \alpha }{ 1 }y+{ \alpha }{ 2 }z=0$ and ${ \beta }{ 1 }y+{ \beta }{ 2 }z=0$ has a non-trivial solution, then

  1. ${ b }^{ 2 }pr={ q }^{ 2 }ac$
  2. $bp{ r }^{ 2 }=qa{ c }^{ 2 }$
  3. $b{ p }^{ 2 }r=q{ a }^{ 2 }c$

  4. None of these
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A Correct answer
Explanation

The system has a non-trivial solution if the determinant of the coefficients is zero: alpha1*beta2 - alpha2*beta1 = 0. This implies alpha1/alpha2 = beta1/beta2. Using properties of roots, this leads to the condition b^2pr = q^2ac.

AI explanation

For the system of 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 the coefficients must be zero, giving alpha_1 * beta_2 - alpha_2 * beta_1 = 0. This means the ratio of the roots for both equations must be equal, so alpha_1 / alpha_2 = beta_1 / beta_2. Substituting Vieta's formulas by dividing -b/a by c/a and -q/p by r/p gives -b/c = -q/r, and cross-multiplying this relationship results in b^2 * p * r = q^2 * a * c.

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