find the zeroes of the quadratic …
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Sia ? 6 years, 7 months ago
p(x) = 3x2 + x - 2 = 0
= 3x2 + 3x - 2x - 2 = 0
= 3x(x + 1) - 2(x + 1) = 0
= (3x - 2)(x + 1) = 0
Hence, zeroes are -1 and {tex}\frac 23{/tex}
Verification:
Sum of zeroes = {tex}- 1 + \frac { 2 } { 3 } = \frac { - 1 } { 3 }{/tex}
Product of zeroes = {tex}( - 1 ) \times \left( \frac { 2 } { 3 } \right) = \frac { -2 } { 3 }{/tex}
Again sum of zeroes = {tex}- \frac { \text { Coeff. of } x } { \text { Coeff. of } x ^ { 2 } } = \frac { - 1 } { 3 }{/tex}
Product of zeroes = {tex}\frac { \text { Constant term } } { \text { Coeff. of } x ^ { 2 } }{/tex} = {tex}\frac{-2}{3}{/tex}
Verified.
0Thank You