Find the zeroes of each of …
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Sia ? 6 years, 5 months ago
The given polynomial is
p(x) = 6x2 -7x -3
Factorize the above quadratic polynomial, we have
6x2 -7x -3 = 6x2 -9x + 2x - 3
= 3x(2x - 3) + 1(2x - 3)
= (3x + 1)(2x - 3)
For p(x) = 0, either 3x + 1 = 0 or 2x - 3 = 0
{tex}\Rightarrow \quad x = \frac { - 1 } { 3 } \text { or } x = \frac { 3 } { 2 }{/tex}
Verification:we have a = 6, b = -7, c = -3
Sum of zeroes = {tex}\frac { - 1 } { 3 } + \frac { 3 } { 2 } = \frac { 7 } { 6 }{/tex}
Also, {tex}\frac { - b } { a } = \frac { - ( - 7 ) } { 6 } = \frac { 7 } { 6 }{/tex}
Now, product of zeroes= {tex}\left( - \frac { 1 } { 3 } \right) \times \frac { 3 } { 2 } = \frac { - 1 } { 2 }{/tex}
Also, {tex}\frac { c } { a } = \frac { - 3 } { 6 } = \frac { - 1 } { 2 } {/tex}
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