Question

find the zero of quadric polynomial X square +7x +10 and verify the relationship between the zeros and coffiecients?Plz reply fast I'm in hurry ​

Answer 1

$\bf{\large{\underline{\underline{\bf{Given\::}}}}}$

The quadratic polynomial x² + 7x + 10

$\bf{\large{\underline{\underline{\bf{To\:find\::}}}}}$

The zeroes and verify the relationship between the zeroes and the coefficient.

$\bf{\large{\underline{\underline{\bf{Explanation\::}}}}}$

We have p(x) = x² + 7x + 10

Zero of the polynomial is p(x) = 0

So;

$\begin{gathered}\longrightarrow\sf{x^{2} +7x+10=0}\\\\\longrightarrow\sf{x^{2} +2x+5x+10=0}\\\\\longrightarrow\sf{x(x+2)+5(x+2)=0}\\\\\longrightarrow\sf{(x+2)(x+5)=0}\\\\\longrightarrow\sf{x+2=0\:\:\:Or\:\:\:x+5=0}\\\\\longrightarrow\sf{\red{x=-2\:\:\:Or\:\:\:x=-5}}\end{gathered}$

∴ The α = -2 and β = -5 are the zeroes of the polynomial.

As the given quadratic polynomial as we compared with ax²+bx+c=0

a = 1

b = 7

c = 10

So;

$\bf{\green{\underline{\underline{\tt{Sum\:of\:the\:zeroes\::}}}}}$

$\begin{gathered}\mapsto\sf{\alpha +\beta=\dfrac{-b}{a} =\dfrac{Coefficient\:of\:x^{2} }{Coefficient\:of\:x}} \\\\\\\mapsto\sf{-2+(-5)=\dfrac{-7}{1} }\\\\\\\mapsto\sf{-2-5=-7}\\\\\\\mapsto\sf{\red{-7=-7}}\end{gathered}$

$\bf{\green{\underline{\underline{\tt{Product\:of\:the\:zeroes\::}}}}}$

$\begin{gathered}\mapsto\sf{\alpha \times \beta=\dfrac{c}{a} =\dfrac{Constant\:term }{Coefficient\:of\:x}} \\\\\\\mapsto\sf{-2\times (-5)=\dfrac{10}{1} }\\\\\\\mapsto\sf{-(-10)=10}\\\\\\\mapsto\sf{\red{10=10}}\end{gathered}$

Thus;

Relationship between zeroes and coefficient is verified .