Find the zeroes of quadratic polynomial …
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Sia ? 4 years, 10 months ago
Here, p(x) = 3x2 - 2.
Now p(x) = 0
{tex}\Rightarrow \quad 3 x ^ { 2 } - 2 = 0 {/tex}
{tex}\Rightarrow 3 x ^ { 2 } = 2{/tex}
{tex}\Rightarrow \quad x ^ { 2 } = \frac { 2 } { 3 } {/tex}
{tex}\Rightarrow x = \pm \sqrt { \frac { 2 } { 3 } }{/tex}
Therefore, zeroes are {tex}\sqrt { \frac { 2 } { 3 } } \text { and } - \sqrt { \frac { 2 } { 3 } }{/tex}.
If p(x) = 3x2 - 2 , then a = 3, b = 0 and c = -2
Now, sum of zeroes = {tex}\sqrt { \frac { 2 } { 3 } } + \left( - \sqrt { \frac { 2 } { 3 } } \right){/tex}= 0 .... (i)
Also, {tex}\frac { - b } { a } = \frac { - 0 } { 3 } = 0 {/tex} ........ (ii)
From (i) and (ii)
Sum of zeroes{tex}= \frac { - b } { a }{/tex}
and product of zeroes = {tex}\sqrt { \frac { 2 } { 3 } } \times - \sqrt { \frac { 2 } { 3 } } = \frac { - 2 } { 3 }{/tex}......... (iii)
Also, {tex}\frac { c } { a } = \frac { - 2 } { 3 } {/tex}.......... (iv)
From (iii) and (iv)
Product of zeroes = {tex}\frac { c } { a }{/tex}
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