Question

Find zeros of the following quadratic polynomial and verify the relationship between the zeros and the coefficients... 3.6x2-x-2​

Answer 1

Answer:

5.2

Step-by-step explanation:

3.6×2-x-2=0

7.2-x-2=0

7.2-2-x=0

5.2--X=0

-X=0-5.2

-X= -5.2

X=5.2 (we can cut the - because they are on both X and 5.2)#

Answer 2

S O L U T I O N :

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

We have the quadratic polynomial : 6x² - x - 2.

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

The zeroes and verify the relationship between zeroes & coefficient.

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

We have p(x) = 6x² - x - 2

Zero of the polynomial p(x) = 0

So;

$\longrightarrow\sf{6x^{2} -x-2=0}\\\\\longrightarrow\sf{6x^{2} -4x+3x-2=0}\\\\\longrightarrow\sf{2x(3x-2)+1(3x-2)=0}\\\\\longrightarrow\sf{(3x-2)(2x+1)=0}\\\\\longrightarrow\sf{3x-2=0\:\:\:Or\:\:\:2x+1=0}\\\\\longrightarrow\sf{3x=2\:\:\:Or\:\:\:2x=-1}\\\\\longrightarrow\sf{x=2/3\:\:\:Or\:\:\:x=-1/2}$

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

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

• a = 6
• b = -1
• c = -2

Now;

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

$\mapsto\sf{\alpha +\beta =\dfrac{-b}{a} =\dfrac{Coefficient\:of\:x}{Coefficient\:of\:x^{2} }} \\\\\\\mapsto\sf{\dfrac{2}{3} +\bigg(-\dfrac{1}{2} \bigg)=\dfrac{-(-1)}{6} }\\\\\\\mapsto\sf{\dfrac{2}{3} -\dfrac{1}{2}=\dfrac{1}{6} }\\\\\\\mapsto\sf{\dfrac{4-3}{6} =\dfrac{1}{6} }\\\\\\\mapsto\bf{\dfrac{1}{6} =\dfrac{1}{6} }$

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

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

Thus;

Relationship between zeroes and coefficient is verified .