Question

If alpha,beta are the roots of the equation 7x^2+ax+2=0 and if beta -alpha =-13\7 ,then find the value of a

Answer 1

Answer:

$a \: = - 15 \: or \: 15$

Step-by-step explanation:

Given---->

$\alpha \: and \: \beta \: are \: the \:zeroes \: of \: the \: equation \\ 7 {x}^{2} + ax + 2 = 0$

$and \: if \: (\beta - \alpha ) = - \dfrac{13}{7}$

To find ---->

$value \: of \: a$

Concept used ----> 1)

$if \: \alpha \: and \: \beta \: are \: roots \: of \: equation \\ a {x}^{2} + bx + c = 0 \: then \\ \alpha + \beta \: = - \dfrac{coefficient \: of \: x}{coefficient \: of \: {x}^{2} } \\ \alpha \beta \: = \dfrac{constant \: term}{coefficient \: of \: {x}^{2} }$

2)

${(a - b)}^{2} \: = {(a + b)}^{2} - 4ab$

Solution----> ATQ,

$7 {x}^{2} + ax + 2 = 0$

$\alpha + \beta = - \dfrac{coefficient \: of \: x}{coefficient \: of \: {x}^{2} }$

$= > \alpha + \beta = - \dfrac{a}{7}$

$= > \alpha \beta = \dfrac{constant \: term}{coefficient \: of \: {x}^{2} }$

$= > \alpha \beta = \dfrac{2}{7}$

$now$

$\beta - \alpha = - \dfrac{13}{7}$

$squaring \: both \: sides \: we \: et$

$= > {( \beta - \alpha )}^{2} = {( - \dfrac{13}{7}) }^{2}$

$= > {( \beta + \alpha )}^{2} - \alpha \beta = \dfrac{169}{49}$

$= > {( - \dfrac{a}{7}) }^{2} - 4( \dfrac{2}{7}) = \dfrac{169}{49}$

$= > \dfrac{ {a}^{2} }{49} - \dfrac{8}{7} = \dfrac{169}{49}$

$= > \dfrac{ {a}^{2} }{49} \: = \dfrac{169}{49} + \dfrac{8}{7}$

$= > \dfrac{ {a}^{2} }{49} = \dfrac{169 + 56}{49}$

$49 \: is \: cancel \: out \: from \: each \: side$

${a}^{2} = 225$

$taking \: square \: root \: of \: both \: sides \:$

$= > a = - 15 \: or \: 15$

Answer 2

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