Question

If alpha and beta are the zeros of the polynomial 3x^2 - 4x - 7 then find the sum of their reciprocals.

Answer 1

✪AnSwEr

${\tt{\red{\underline{\large{Given}}}}}$

• A polynomial
• 3x²-4x-7

${\sf{\green{\underline{\large{To\:Find}}}}}$

• sum of thier reciprocal
• Relationship between cofficient

${\sf{\pink{\underline{\Large{Explanation}}}}}$

Let the zeroes of the polynomial be$\tt\alpha{and}\beta$

Then,

$\rightarrow\tt\alpha{+}\beta{=}\frac{4}{3}$

&

$\rightarrow\tt\alpha{\times}\beta{=}\frac{-7}{3}$

1/a + 1/b

=>{a+b}/ab

=>4/3 × 3/-7

=>-4/7

☞

Additional Information

$\rightarrow\tt\alpha{+}\beta{=}\dfrac{4}{3}$

$\rightarrow\tt\alpha{+}\beta{=}\dfrac{Cofficient\:of\:X}{Cofficient\:of\:x^2}=$

&

$\rightarrow\tt\alpha{\times}\beta{=}\dfrac{-7}{3}$

$\rightarrow\tt{\large\alpha{\times}\beta{=}\dfrac{Constant\:term}{Cofficient\:of\:x^2}}$

Hence,relation verified

Answer 2

$\bigstar$ Given :

• If α and β are the zeros of the polynomial 3x² - 4x - 7

$\bigstar$ To Find :

• The sum of their reciprocals

$\bigstar$ Solution :

Compare given equation 3x²-4x-7 with ax²+bx+c , we get ,

• a = 3 , b = - 4 , c - 7

Sum of zeroes ,

$\alpha +\beta =\frac{-b}{a}\\\\ \implies \alpha +\beta =\frac{-(-4)}{3} \\\\\implies\bold{ \alpha +\beta =\frac{4}{3} }$

Product of zeroes ,

$\alpha \beta =\frac{c}{a}\\\\ \implies\bold{ \alpha \beta =\frac{-7}{3} }$

So , Sum of reciprocals ,

$\frac{1}{\alpha } +\frac{1}{\beta } \\\\\implies \frac{\alpha+\beta }{\alpha \beta } \\\\\implies \frac{\frac{4}{3} }{\frac{-7}{3} } \\\\\implies \bold{\bf{\blue{-\frac{4}{7} }}}$