Question

Find the zeros of each of the following quadratic polynomial and verify the relationship between the zeros and their coefficients :
(i) $f\left(x\right)=x^2-2x-8$

(ii) $g\left(x\right)=4s^2-4s+1$

(iii) $f\left(x\right)=x^2-(\sqrt{3}+1)x+\sqrt{3}$

(iv) $x^2-3-7x$

(v) $p\left(x\right)=x^2+2\sqrt{2}x-6$

(vi) $q\left(x\right)=\sqrt{3}{x}^2+10x+7\sqrt{3}$
(vii) $g\left(x\right)=a\left(x^2+1\right)-x\left(a^2+1\right)$

Answer 1

$a{x}^2+bx+c=0\Rightarrow \alpha +\beta =-\frac{b}{a},\alpha \beta =\frac{c}{a}$

(i)

$x^2-2x-8=0$

$\Rightarrow a=1,b=-2,c=-8$

$x^2-2x-8=0$

$(x-4)(x+2)=0$

$\alpha =-2,\beta =4$

$\alpha +\beta =-\frac{b}{a}\to -2+4=-\frac{-2}{1}\Rightarrow 2=2$

$\alpha \beta =\frac{c}{a}\to (-2)\left(4\right)=\frac{-8}{1}\Rightarrow -8=-8$

Hence the relationship between zeros and coefficients is verified.

(ii)

$4s^2-4s+1=0$

$\Rightarrow a=4,b=-4,c=1$

$4s^2-4s+1=0$

$(2s-1)(2s-1)=0$

$\alpha =\frac{1}{2},\beta =\frac{1}{2}$

$\alpha +\beta =-\frac{b}{a}\to \frac{1}{2}+\frac{1}{2}=-\frac{-4}{4}\Rightarrow 1=1$

$\alpha \beta =\frac{c}{a}\to \frac{1}{2}\times \frac{1}{2}=\frac{1}{4}\Rightarrow \frac{1}{4}=\frac{1}{4}$

Hence the relationship between zeros and coefficients is verified.

(iii)

$x^2-(\sqrt{3}+1)x+\sqrt{3}=0$

$\Rightarrow a=1,b=-(\sqrt{3}+1),c=\sqrt{3}$

$x^2-(\sqrt{3}+1)x+\sqrt{3}=0$

$x=\frac{-(-\sqrt{3}-1)\pm \sqrt{{(-\sqrt{3}-1)}^2-4\cdot 1\sqrt{3}}}{2\cdot 1}=\sqrt{3},1$

$\alpha =\sqrt{3},\beta =1$

$\alpha +\beta =-\frac{b}{a}\to \sqrt{3}+1=-\frac{-(\sqrt{3}+1)}{1}\Rightarrow LHS=RHS$

$\alpha \beta =\frac{c}{a}\to 1\times \sqrt{3}=\frac{\sqrt{3}}{1}\Rightarrow LHS=RHS$

Hence the relationship between zeros and coefficients is verified.

(iv)

$x^2-7x-3=0$

$\Rightarrow a=1,b=-7,c=-3$

$x^2-7x-3=0$

$x=\frac{-(-7)\pm \sqrt{{(-7)}^2-4\cdot 1(-3)}}{2\cdot 1}:\frac{7\pm \sqrt{61}}{2}$

$\alpha =\frac{7+\sqrt{61}}{2},\beta =\frac{7-\sqrt{61}}{2}$

$\alpha +\beta =-\frac{b}{a}\to \frac{7+\sqrt{61}}{2}+\frac{7-\sqrt{61}}{2}=-\frac{-7}{1}\Rightarrow LHS=RHS$

$\alpha \beta =\frac{c}{a}\to \frac{7+\sqrt{61}}{2}\times \frac{7-\sqrt{61}}{2}=\frac{-3}{1}\Rightarrow LHS=RHS$

Hence the relationship between zeros and coefficients is verified.

(v)

$x^2+2\sqrt{2}x-6=0$

$\Rightarrow a=1,b=2\sqrt{2},c=-6$

$x^2+2\sqrt{2}x-6=0$

$x=\frac{-2\sqrt{2}\pm \sqrt{{\left(2\sqrt{2}\right)}^2-4\cdot 1(-6)}}{2\cdot 1}=\sqrt{2},-3\sqrt{2}$

$\alpha =\sqrt{2},\beta =-3\sqrt{2}$

$\alpha +\beta =-\frac{b}{a}\to \sqrt{2}-3\sqrt{2}=-\frac{2\sqrt{2}}{1}\Rightarrow LHS=RHS$

$\alpha \beta =\frac{c}{a}\to \left(\sqrt{2}\right)(-3\sqrt{2})=\frac{-6}{1}\Rightarrow LHS=RHS$

Hence the relationship between zeros and coefficients is verified.

(vi)

$\sqrt{3}{x}^2+10x+7\sqrt{3}=0$

$\Rightarrow a=\sqrt{3},b=10,c=7\sqrt{3}$

$\sqrt{3}{x}^2+10x+7\sqrt{3}=0$

$x=\frac{-10\pm \sqrt{{10}^2-4\sqrt{3}7\sqrt{3}}}{2\sqrt{3}}=-\sqrt{3},-\frac{7}{\sqrt{3}}$

$\alpha =-\sqrt{3},\beta =-\frac{7}{\sqrt{3}}$

$\alpha +\beta =-\frac{b}{a}\to -\sqrt{3}-\frac{7}{\sqrt{3}}=-\frac{10}{\sqrt{3}}\Rightarrow LHS=RHS$

$\alpha \beta =\frac{c}{a}\to (-\sqrt{3})(-\frac{7}{\sqrt{3}})=\frac{7\sqrt{3}}{\sqrt{3}}\Rightarrow LHS=RHS$

Hence the relationship between zeros and coefficients is verified.

(vii)

$a(x^2+1)-x(a^2+1)=0$

$a{x}^2-(a^2+1)x+a=0$

$\Rightarrow a=a,b=-(a^2+1),c=a$

$a{x}^2-(a^2+1)x+a=0$

$x=\frac{-(-a^2-1)\pm \sqrt{{(-a^2-1)}^2-4aa}}{2a}=a,\frac{1}{a}$

$\alpha =a,\beta =\frac{1}{a}$

$\alpha +\beta =-\frac{b}{a}\to a+\frac{1}{a}=-\frac{-(a^2+1)}{a}\Rightarrow LHS=RHS$

$\alpha \beta =\frac{c}{a}\to a\times \frac{1}{a}=\frac{a}{a}\Rightarrow LHS=RHS$

Hence the relationship between zeros and coefficients is verified.