Question

Find the zeros of each of the following quadratic polynomials and verify the relationship between the zeros and their coefficients :

(i) $f\left(x\right)=x^2-2x-8$

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

(iii) $h\left(t\right)=t^2-15$

(iv) $6x^2-3-7x$

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

(vi) $q\left(x\right)=\sqrt{3x^2}+10x+\sqrt[7]{73}$

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

(viii) $g\left(x\right)=a(x^2+1)-x(a^2+1)$

Answer 1

i) $f\left(x\right)=x^2-2x-8$

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

Zeroes: $-2,4$

Sum of zeroes: $-2+4=2$

Product of zeroes:$-2\left(4\right)=-8$

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

$=(2s-1{)}^2$

Zeroes:$\frac{1}{2},\frac{1}{2}$

Sum of zeroes : $\frac{1}{2}+\frac{1}{2}=\frac{-b}{a}=-\frac{(-4)}{4}=1$

Product of zeroes:$\frac{1}{2}\cdot \frac{1}{2}=\frac{c}{a}=\frac{1}{4}$

ii)$h\left(t\right)=t^2-15=(t-\sqrt{1}5)(t+\sqrt{1}5)$

Sum of zeroes: $\sqrt{1}5+(-\sqrt{1}5)=0$

Product of zeroes: $\left(\sqrt{1}5\right)(-\sqrt{1}5)=-15=\frac{c}{a}$

iv) $6x^2-3-7x$

Zeroes : $\frac{3}{2},\frac{-1}{3}$

Sum of zeroes:$\frac{3}{2}-\frac{1}{3}=\frac{7}{6}=\frac{-b}{a}$

Product of zeroes: $\frac{3}{2}\frac{-1}{3}=\frac{-3}{6}=\frac{c}{a}$

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

$=(x-\sqrt{6-2\sqrt{2}})(x+\sqrt{6-2\sqrt{2}})$

Zeroes: $-\sqrt{6-2\sqrt{2}},\sqrt{6-2\sqrt{2}}$

Sum of zeroes:$\sqrt{6-2\sqrt{2}}-\sqrt{6-2\sqrt{2}}=0=\frac{-b}{a}$

Product of zeroes: $\left(\sqrt{6-2\sqrt{2}}\right)(-\sqrt{6-2\sqrt{2}})=-6+2\sqrt{2}=\frac{c}{a}$

vi) $q\left(x\right)=\sqrt{3}{x}^2+10x+7\sqrt{3}$

Zeroes:$-\sqrt{3},\frac{-7}{\sqrt{3}}$

Sum of zeroes: $-(\sqrt{3}+\frac{7}{\sqrt{3}})=\frac{-10}{\sqrt{3}}=\frac{-b}{a}$

Product of zeroes: $(-\sqrt{3})(-\frac{7}{\sqrt{3}})=7=\frac{7\sqrt{3}}{\sqrt{3}}=\frac{c}{a}$

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

Zeroes: $1,\sqrt{3}$

Sum of zeroes: $1+\sqrt{3}=\frac{c}{a}$

Product of zeroes: $1\times \sqrt{3}=\sqrt{3}=\frac{c}{a}$

viii)$g\left(x\right)=a(x^2+1)-x(a^2+1)=a{x}^2-(a^2+1)x+a$

Zeroes: $a,\frac{1}{a}$

Sum of zeroes: $a+\frac{1}{a}=\frac{(a^2+1)}{a}=\frac{b^{{}^{\prime }}}{c^{{}^{\prime }}}$

Product of zeroes: $a\times \frac{1}{a}=1=\frac{a}{a}=\frac{c^{{}^{\prime }}}{a^{{}^{\prime }}}$