Question

6x^2+7x+1
find the zeroes of the following quadratic polynomial and verify the relationship between the zeroes and coefficient​

Answer 1

Answer:

Hope it will woek for you

Answer 2

S O L U T I O N :

We have p(x) = 6x² + 7x + 1

Zero of the polynomial p(x) = 0

Using quadratic formula :

As we know that given polynomial compared with ax² + bx + c;

• a = 6
• b = 7
• c = 1

$\longrightarrow\tt{x=\dfrac{-b\pm\sqrt{b^{2}-4ac } }{2a} }\\\\\\\longrightarrow\tt{x=\dfrac{-7\pm\sqrt{(7)^{2} -4\times 6\times 1} }{2\times 6} }\\\\\\\longrightarrow\tt{x=\dfrac{-7\pm\sqrt{49-24} }{12} }\\\\\\\longrightarrow\tt{x=\dfrac{-7\pm\sqrt{25} }{12} }\\\\\\\longrightarrow\tt{x=\dfrac{-7\pm5}{12} }\\\\\\\longrightarrow\tt{x=\dfrac{-7+5}{12} \:\:Or\:\:x=\dfrac{-7-5}{12} }\\\\\\\longrightarrow\tt{x=\cancel{\dfrac{-2}{12}} \:\:Or\:\:x=\cancel{\dfrac{-12}{12} }}$

$\longrightarrow\bf{x=-\dfrac{1}{6} \:\:\:Or\:\:\:x=-1}$

∴ The α = -1/6 and β = -1 are the zeroes of the polynomial.

Now;

$\underline{\mathcal{SUM\:OF\:THE\:ZEROES\::}}$

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

$\underline{\mathcal{PRODUCT\:OF\:THE\:ZEROES\::}}$

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

Thus;

Relationship between zeroes and coefficient is verified .