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Zeroes of polynomial of 5x2-4-8x

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Zeroes of polynomial of 5x2-4-8x

Posted by Yathath Yadav 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

p(x) = 5x² + 8x - 4 = 0
= 5x² + 10x - 2x - 4 = 0
= 5x(x + 2) - 2(x + 2) = 0
= (x + 2)(5x - 2) = 0
Hence, zeroes are -2 and {tex}\frac 25{/tex}
Verification:
Sum of zeroes = {tex}- 2 + \frac { 2 } { 5 } = \frac { - 8 } { 5 }{/tex}
Product of zeroes = {tex}( - 2 ) \times \left( \frac { 2 } { 5 } \right) = \frac { - 4 } { 5 }{/tex}
Again sum of zeroes = {tex}- \frac { \text { Coeff. of } x } { \text { Coeff. of } x ^ { 2 } } = \frac { - 8 } { 5 }{/tex}
Product of zeroes = {tex}\frac { \text { Constant term } } { \text { Coeff. of } x ^ { 2 } }{/tex} = {tex}\frac{-4}{5}{/tex}
Verified.

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