Question

1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeros and the coefficients.

(i) x^2-2x-8

(ii) 4s^2-4s+1

(iii) t^2-15

Answer 1

Refer to the attachments.

Answer 2

Solution :

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

The quadratic polynomial :

• x² - 2x - 8
• 4s² - 4s + 1
• t² - 15

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

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

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

We have p(x) = x² - 2x - 8.

Zero of the polynomial p(x) = 0

So;

$\longrightarrow\rm{x^{2} -2x-8=0}\\\\\longrightarrow\rm{x^{2} -4x+2x-8=0}\\\\\longrightarrow\rm{x(x-4)+2(x-4)=0}\\\\\longrightarrow\rm{(x-4)(x+2)=0}\\\\\longrightarrow\rm{x-4=0\:\:\:Or\:\:\:x+2=0}\\\\\longrightarrow\rm{\orange{x=4\:\:\:Or\:\:\:x=-2}}$

∴ The α = 4 and β = -2 are the zeroes of the polynomial.  

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

• a = 1
• b = -2
• c = -8

Now;

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

$\mapsto\rm{\alpha +\beta =\dfrac{-b}{a} =\dfrac{Coefficient\:of\:x}{Coefficient\:of\:x^{2} } }\\\\\\\mapsto\rm{4+(-2)=\dfrac{-(-2)}{1} }\\\\\\\mapsto\rm{4-2=2}\\\\\\\mapsto\rm{\orange{2=2}}$

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

$\mapsto\rm{\alpha \times \beta =\dfrac{c}{a} =\dfrac{Constant\:term}{Coefficient\:of\:x^{2} } }\\\\\\\mapsto\rm{4\times (-2)=\dfrac{-8}{1} }\\\\\\\mapsto\rm{\orange{-8=-8}}$

Thus;

Relationship between zeroes and coefficient is verified .

$\star\:\boxed{\large{\bf{4s^{2} -4s+1}}}}$

We have p(x) = 4s² - 4s + 1.

Zero of the polynomial p(x) = 0

So;

$\longrightarrow\rm{4s^{2} -4s+1=0}\\\\\longrightarrow\rm{4s^{2} -2s-2s+1=0}\\\\\longrightarrow\rm{2s(2s-1)-1(2s-1)=0}\\\\\longrightarrow\rm{(2s-1)(2s-1)=0}\\\\\longrightarrow\rm{2s-1=0\:\:\:Or\:\:\:2s-1=0}\\\\\longrightarrow\rm{2s=1\:\:\:Or\:\:\:2s=1}\\\\\longrightarrow\rm{\orange{s=1/2\:\:\:Or\:\:\:s=1/2}}$

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

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

• a = 4
• b = -4
• c = 1

Now;

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

$\mapsto\rm{\alpha +\beta =\dfrac{-b}{a} =\dfrac{Coefficient\:of\:x}{Coefficient\:of\:x^{2} } }\\\\\\\mapsto\rm{\dfrac{1}{2} +\dfrac{1}{2} =\dfrac{-(-4)}{4} }\\\\\\\mapsto\rm{\dfrac{1+1}{2} =\dfrac{4}{4} }\\\\\\\mapsto\rm{\cancel{\dfrac{2}{2}} =\cancel{\dfrac{4}{4}} }\\\\\\\mapsto\rm{\orange{1=1}}$

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

$\mapsto\rm{\alpha \times \beta =\dfrac{c}{a} =\dfrac{Constant\:term}{Coefficient\:of\:x^{2} } }\\\\\\\mapsto\rm{\dfrac{1}{2} \times \dfrac{1}{2} =\dfrac{1}{4} }\\\\\\\mapsto\rm{\orange{\dfrac{1}{4} =\dfrac{1}{4} }}$

Thus;

Relationship between zeroes and coefficient is verified .

$\star\:\boxed{\large{\bf{t^{2} -15}}}}$

We have p(x) = t² - 15

Zero of the polynomial p(x) = 0

So;

$\longrightarrow\rm{t^{2} -15=0}\\\\\longrightarrow\rm{t^{2} =15}\\\\\longrightarrow\rm{\orange{t=\pm\sqrt{15} }}$

∴ The α = √15 and β = -√15 are the zeroes of the polynomial.  

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

• a = 1
• b = 0
• c = -15

Now;

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

$\mapsto\rm{\alpha +\beta =\dfrac{-b}{a} =\dfrac{Coefficient\:of\:x}{Coefficient\:of\:x^{2} } }\\\\\\\mapsto\rm{\sqrt{15} +(-\sqrt{15} )=\dfrac{0}{1} }\\\\\\\mapsto\rm{\sqrt{15} -\sqrt{15} =0}\\\\\\\mapsto\rm{\orange{0=0}}$

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

$\mapsto\rm{\alpha \times \beta =\dfrac{c}{a} =\dfrac{Constant\:term}{Coefficient\:of\:x^{2} } }\\\\\\\mapsto\rm{\sqrt{15} \times (-\sqrt{15}) =\dfrac{-15}{1} }\\\\\\\mapsto\rm{-\sqrt{15\times 15} =-15}\\\\\\\mapsto\rm{\orange{-15=-15}}$

Thus;

Relationship between zeroes and coefficient is verified .