Question

if alpha, beta are the roots of the equation ax square + bx + c equal to zero and alpha +k, beta + K are the roots of the equation px square + q x + r are equal to zero then K equal to​

Answer 1

$\Huge{\underline{\underline{\mathfrak{Question \colon }}}}$

If $\alpha , \beta$ are the roots of the equation ax² + bx + c = 0 and, $\alpha + K, \beta + K$ are the roots of the equation px² + qx + r = 0, then K equal to

$\Huge{\underline{\underline{\mathfrak{Answer \colon }}}}$

Given that,

$\alpha \ and \ \beta$ are the zeros of the equation ax² + bx + c = 0 and, $\alpha + K \ and \ \beta + K$ are the zeros of the equation px² + qx + r

To find

Value of K

Sum of Zeros

$\large{\boxed{ \sf{ \alpha + \beta = \: - \dfrac{x \: coefficient}{x {}^{2} \: coefficient } }}}$

Consider equation ax² + bx + c = 0,

$\large{\sf{ \alpha + \beta } = - \dfrac{ b}{a} \: \: \: - - - - - - - - (1) }$

Consider equation px² + qx + r = 0,

$\large{ \sf{( \alpha + K) + ( \beta + K) = - \dfrac{q}{p} }} \\ \\ \large{ \implies \: \sf{( \alpha + \beta ) + 2K = - \frac{ q}{p} }}$

From equation (1),

$\large{ \sf{ - \dfrac{b}{a} + 2K = - \dfrac{p}{q} }} \\ \\ \large{ \implies \: \sf{2K = \dfrac{bq - ap}{aq} }} \\ \\ \huge{ \implies \: \boxed{ \boxed{ \sf{ \green{K = \frac{bq - ap}{aq} }}}}}$

Answer 2

Answer:

$\huge{\underline{\underline{\mathfrak{Answer\::}}}}$

K = (bp-aq)/2ap

___________________________

$(\alpha + \beta )are \: the \: roots \: of = \\ ax {}^{2} + bx + c \\ \\ (\alpha + k)( \beta + k)are \: the \: roots \: of = \\ px {}^{2} + qx + r \\consider \: ax {}^{2} + bx + c = 0 \\ (\alpha + \beta ) = - \frac{co \: efficient \: of \: x}{co \: efficient \: of \: x {}^{2} } \\ (\alpha + \beta ) = - \frac{b}{a} \\ \\ consider \: px {}^{2} + qx + r = 0 \\ (\alpha + k) (\beta + k) = - \frac{co \: efficient \: of \: x}{co \: efficient \: of \: x {}^{2} } \\ (\alpha + k)( \beta + k) = - \frac{q}{p} \\ (\alpha + \beta ) + 2k = - \frac{q}{p} \\ - \frac{b}{a} +2 k = - \frac{q}{p} \\ 2k = \frac{b}{a} - \frac{q}{p} \\ 2k = \frac{bp - aq}{ap} \\ k = \frac{bp - aq}{2ap}$

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