Question

if alpha beta are the zero of polynominal p(x) 4x ² + 3x +7 then
$1 \div \alpha + 1 \div \beta$
is equal to​

Answer 1

Answer:

3/7

Step-by-step explanation:

4x²+3x+7=0

let the zeros of the equation be a & b

so a+b=3/4..........eq1

a*b=7/4...........eq2

using eq2

1/a=4b/7

1/b=4a/7

as in question,

(1/a)+(1/b)=(4b/7)+(4a/7)

=(a+b) * 4/7 put (a+b)=3

=12/7

is your answer

Answer 2

$\bold{\huge{\underline{\underline{\rm{ Given :}}}}}$

$\alpha \: and \: \beta$ Are the zeros of the Polynomial $4 {x}^{2} + 3x + 7$

$\bold{\huge{\underline{\underline{\rm{ To\:Find :}}}}}$

$\frac{1}{ \alpha } + \frac{1}{ \beta } =$

$\rule{200}{1}$

$\</em></strong><strong>r</strong><strong><em>e</em></strong><strong><em>d</em></strong><strong><em>{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

First find the zeros of given polynomial-

$4 {x}^{2} + 3x + 7$

Comparing this polynomial by ax² + bx + c = 0

=> a = 4

=> b = 3

=> c + 7

Formula to find Zeros -

• $x = \frac{ - b + \sqrt{ {b}^{2} - 4ac} }{2a}$

$x = \frac{ - 3 + \sqrt{( {3)}^{2} - 4 \times 4 \times 7 } }{2 \times 4} \\ x = \frac{ - 3 + \sqrt{9 - 112} }{8} \\ x = \frac{ - 3 + \sqrt{ - 103} }{8}$

so ,

$\alpha = \frac{ - 3 + \sqrt{ - 103} }{8} \\ \beta = \frac{ - 3 - \sqrt{ - 103} }{8}$

$\rule{200}{1}$

$\frac{1}{ \alpha } + \frac{1}{ \beta } =$

$= \frac{1}{ \frac{ - 3 + \sqrt{ - 103} }{8} } + \frac{1}{ \frac{ - 3 - \sqrt{ - 103} }{8} } \\ \\ = \frac{8}{ \sqrt{ - 103} - 3 } - \frac{8}{ \sqrt{ - 103 } + 3 } \\ \\ = \frac{8( \sqrt{ - 103} + 3) - 8( \sqrt{ - 103} - 3) }{( \sqrt{ - 103} - 3)( \sqrt{ - 103} + 3) } \\ \\ = \frac{8( \sqrt{ - 103} + 3 - \sqrt{ - 103 } + 3) }{( { \sqrt{ - 103} )}^{2} - ( {3})^{2} } \\ \\ = \frac{8(6)}{ - 103 - 9} \\ \\ = \frac{48}{ - 12} \\ \\ = - 4$

$\rule{200}{1}$

So,

$\frac{1}{ \alpha } + \frac{1}{ \beta } = - 4$

$\rule{200}{1}$

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