Question

find the zeros of 6x'-3-7x quardic polynomials and verify the relationship betweeen the zeros and the coeffients​

Answer 1

Answer:

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Answer 2

Solution :

We have p(x) = 6x² - 7x - 3

Zero of the polynomial p(x) = 0

So;

$\longrightarrow\tt{6x^{2} -7x-3=0}\\\\\longrightarrow\tt{6x^{2} -9x+2x-3=0}\\\\\longrightarrow\tt{3x(2x-3)+1(2x-3)=0}\\\\\longrightarrow\tt{(2x-3)(3x+1)=0}\\\\\longrightarrow\tt{2x-3=0\:\:Or\:\:3x+1=0}\\\\\longrightarrow\tt{2x=3\:\:\:Or\:\:\:3x=-1}\\\\\longrightarrow\bf{x=3/2\:\:\:Or\:\:\:x=-1/3}$

∴ The α = 3/2 and β = -1/3 are the zeroes of the polynomial.  

As the given quadratic polynomial as we compared with ax² + bx + c;

• a = 6
• b = -7
• c = -3

Now;

$\underline{\boldsymbol{Sum\:of\:the\:zeroes\::}}}$

$\mapsto\tt{\alpha +\beta =\dfrac{-b}{a} =\dfrac{Coefficient\:of\:x}{Coefficient\:of\:x^{2} } }\\\\\\\mapsto\tt{\dfrac{3}{2} +\bigg(-\dfrac{1}{3} \bigg)=\dfrac{-(-7)}{6} }\\\\\\\mapsto\tt{\dfrac{3}{2}-\dfrac{1}{3} =\dfrac{7}{6} }\\\\\\\mapsto\tt{\dfrac{9-2}{6} =\dfrac{7}{6} }\\\\\\\mapsto\tt{\dfrac{7}{6} =\dfrac{7}{6} }$

$\underline{\boldsymbol{Product\:of\:the\:zeroes\::}}}$

$\mapsto\tt{\alpha \times \beta =\dfrac{c}{a} =\dfrac{Constant\:term}{Coefficient\:of\:x^{2} } }\\\\\\\mapsto\tt{\dfrac{3}{2} \times \bigg(-\dfrac{1}{3} \bigg)=\dfrac{-3}{6} }\\\\\\\mapsto\tt{\dfrac{-3}{6}=\dfrac{-3}{6} }$

Thus;

Relationship between zeroes and coefficient is verified .