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

Solution:

(i) We have,

f(x) = x2 − 2x − 8

f(x) = x2 + 2x − 4x − 8

f(x) = x (x + 2) − 4(x + 2)

f(x) = (x + 2) (x − 4)

The zeros of f(x) are given by 

f(x) = 0

x2 − 2x − 8 = 0

(x + 2) (x − 4) = 0

x + 2 = 0

x = −2

Or

x − 4 = 0

x = 4

Thus, the zeros of f(x) = x2 − 2x − 8 are α = −2 and β = 4

Now,

Sum of the zeros $=\alpha+\beta$

$=(-2)+4$

$=-2+4$

$=2$

and

$=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$

 

$=-\left(\frac{-2}{1}\right)$

= 2

Therefore, sum of the zeros $=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$

Product of the zeros $=\alpha \beta$

$=-2 \times 4$

$=-8$

and

$=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$

$=\frac{-8}{1}$

$=-8$

Therefore,

Product of the zeros $=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$

Hence, the relation-ship between the zeros and coefficient are verified.

(ii) Given $g(s)=4 s^{2}-4 s+1$

When have,

g(s) = 4s2 − 4s + 1

g(s) = 4s2 − 2s − 2s + 1

g(s) = 2s (2s − 1) − 1(2s − 1)

g(s) = (2s − 1) (2s − 1)

The zeros of g(s) are given by 

$g(s)=0$

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

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

$(2 s-1)=0$

$2 s=+1$

$s=\frac{+1}{2}$

Or

$(2 s-1)=0$

$2 s=1$

$s=\frac{1}{2}$

Thus, the zeros of $g(x)=4 s^{2}-4 s+1$ are

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

Now, sum of the zeros $=\alpha+\beta$

$=\frac{1}{2}+\frac{1}{2}$

$=\frac{1+1}{2}$

$=\frac{2}{2}$

= 1

and

$\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$

$=-\frac{-4}{4}$

= 1

Therefore, sum of the zeros $=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$

$=\frac{1}{2} \times \frac{1}{2}$

$=\frac{1}{4}$

and $=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$

$=\frac{1}{4}$

Therefore, the product of the zeros $=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$

Hence, the relation-ship between the zeros and coefficient are verified.

(iii) Given $h(t)=t^{2}-15$

We have,

$h(t)=t^{2}-15$

$h(t)=(t)^{2}-(\sqrt{15})^{2}$

$h(t)=(t+\sqrt{15})(t-\sqrt{15})$

The zeros of $h(t)$ are given by

$h(t)=0$

$(t-\sqrt{15})(t+\sqrt{15})=0$

$(t-\sqrt{15})=0$

$t=\sqrt{15}$

Or

$(t+\sqrt{15})=0$

$t=-\sqrt{15}$

Hence, the zeros of $h(t)$ are $\alpha=\sqrt{15}$ and $\beta=-\sqrt{15}$.

Now,

Sum of the zeros $=\alpha+\beta$

$=\sqrt{15}+(-\sqrt{15})$

$=\sqrt{15}-\sqrt{15}$

$=0$

and $=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$

$=\frac{0}{1}$

= 0

Therefore, sum of the zeros $=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$

also,

Product of the zeros $=\alpha \beta$

$=\sqrt{15} \times-\sqrt{15}$

$=-15$

and,

Constant term

 

$\overline{\text { Coefficient of } x^{2}}$

$=\frac{-15}{1}$

$=-15$

Therefore, the product of the zeros $=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$

Hence, The relationship between the zeros and coefficient are verified.

(iv) Given $f(x)=6 x^{2}-3-7 x$

We have, $f(x)=6 x^{2}-7 x-3$

$f(x)=6 x^{2}-9 x+2 x-3$

$f(x)=3 x(2 x-3)+1(2 x-3)$

$f(x)=(3 x+1)(2 x-3)$

The zeros of $f(x)$ are given by

$f(x)=0$

$6 x^{2}-7 x-3=0$

$(3 x+1)(2 x-3)=0$

$3 x=-1$

$x=\frac{-1}{3}$

Or

$2 x-3=0$

$2 x=3$

$x=\frac{3}{2}$

Thus, the zeros of $f(x)=6 x^{2}-7 x-3$ are $\alpha=\frac{-1}{3}$ and $\beta=\frac{3}{2}$.

Now,

Sum of the zeros $=\alpha+\beta$

$=\frac{-1}{3}+\frac{3}{2}$

$=\frac{-1 \times 2}{3 \times 2}+\frac{3 \times 3}{2 \times 3}$

$=\frac{-2}{6}+\frac{9}{6}$

$=\frac{-2+9}{6}$

$=\frac{7}{6}$

and, $=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$

$=\frac{-(-7)}{6}$

$=\frac{7}{6}$

Therefore, sum of the zeros $=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$

Product of the zeros = α × β

$=\frac{-1}{2}$

and, $=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$

$=\frac{-3}{6}$

$=\frac{-1}{2}$

Product of zeros $=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$

Hence, the relation between the zeros and its coefficient are verified.

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

We have,

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

$p(x)=x^{2}+3 \sqrt{2} x-\sqrt{2} x-6$

$p(x)=x(x+3 \sqrt{2})-\sqrt{2}(x+3 \sqrt{2})$

$p(x)=(x-\sqrt{2})(x+3 \sqrt{2})$

The zeros of $p(x)$ are given by

$p(x)=0$

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

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

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

$(x-\sqrt{2})=0$

$x=\sqrt{2}$

Or

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

$x=-3 \sqrt{2}$

Thus, The zeros of $p(x)=x^{2}+2 \sqrt{2} x-6$ are $\alpha=\sqrt{2}$ and $\beta=-3 \sqrt{2}$

Now,

Sum of the zeros $=\alpha+\beta$

$=\sqrt{2}-3 \sqrt{2}$

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

$=\sqrt{2}(-2)$

$=-2 \sqrt{2}$

and

$=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$

$=\frac{-2 \sqrt{2}}{1}$

$=-2 \sqrt{2}$

Therefore, Sum of the zeros $=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$

Product of the zeros $=\alpha \times \beta$

$=\sqrt{2} \times-3 \sqrt{2}$

$=-3 \times 2$

$=-6$

and

$\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$

$=\frac{-6}{1}$

$=-6$

Therefore, The product of the zeros $=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$

Hence, the relation-ship between the zeros and coefficient are verified.

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

We have, $q(x)=\sqrt{3} x^{2}+10 x+7 \sqrt{3}$

$q(x)=\sqrt{3} x^{2}+3 x+7 x+7 \sqrt{3}$

$q(x)=\sqrt{3} x^{2}+\sqrt{3} \times \sqrt{3} \times x+7 x+7 \sqrt{3}$

$q(x)=\sqrt{3} x(x+\sqrt{3})+7(x+\sqrt{3})$

$q(x)=(x+\sqrt{3})(\sqrt{3} x+7)$

The zeros of g(x) are given by 

$g(x)=0$

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

$(x+\sqrt{3})(\sqrt{3} x+7)=0$

$x+\sqrt{3}=0$

$x=-\sqrt{3}$

Or

$\sqrt{3}+7=0$

$\sqrt{3} x=-7$

$x=\frac{-7}{\sqrt{3}}$

Thus, the zeros of $q(x)=\sqrt{3} x^{2}+10 x+7 \sqrt{3}$ are $\alpha=-\sqrt{3}$ and $\beta=\frac{-7}{\sqrt{3}}$.

Now, 

Sum of the zeros $=\alpha+\beta$

$=-\sqrt{3}+\frac{-7}{\sqrt{3}}$

$=\frac{-\sqrt{3} \times \sqrt{3}}{1 \times \sqrt{3}}+\frac{-7}{\sqrt{3}}$

$=\frac{-3}{\sqrt{3}}+\frac{-7}{\sqrt{3}}$

$=\frac{-3-7}{\sqrt{3}}$

$=\frac{-10}{\sqrt{3}}$

and $=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$

$=\frac{-(+10)}{\sqrt{3}}$

$=\frac{-10}{\sqrt{3}}$

Therefore, sum of the zeros $=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$

Product of zeros $=\alpha \times \beta$

$=+7$

and $=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$

$=7$

Therefore, the product of the zeros $=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$

Hence, the relation-ship between the zeros and coefficient are verified.

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

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

$f(x)=x(x-\sqrt{3})-1(x-\sqrt{3})$

$f(x)=(x-1)(x-\sqrt{3})$

The zeros of ƒ(x) are given by 

$f(x)=0$

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

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

$(x-1)=0$

$x=0+1$

$x=1$

Or

$x-\sqrt{3}=0$

$x=0+\sqrt{3}$

$x=\sqrt{3}$

Thus, the zeros of $x^{2}-(\sqrt{3}+1) x+\sqrt{3}$ are $\alpha=1$ and $\beta=\sqrt{3}$

Now,

Sum of zeros $=\alpha+\beta$

$=1+\sqrt{3}$

$=1+\sqrt{3}$

and

$=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$

$=-\frac{-(\sqrt{3}+1)}{1}$

$=\frac{+(\sqrt{3}+1)}{1}$

Therefore, sum of the zeros $=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$

Product of the zeros $=\alpha \beta$

$=1 \times \sqrt{3}$

$=\sqrt{3}$

And

$=\frac{\text { Constant term }}{\text { Con } 2}$

$=\frac{\sqrt{3}}{1}$

$=\sqrt{3}$

Product of zeros $=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$

Hence, the relation-ship between the zeros and coefficient are verified.

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

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

$g(x)=x a(x-a)-1(x-a)$

$g(x)=(x a-1)(x-a)$

The zeros of g(x) are given by

$g(x)=0$

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

$x a-1=0$

$x a=1$

$x=\frac{1}{a}$

Or

$x-a=0$

$x=a$

Thus, the zeros of $g(x)=a x^{2}-\left(a^{2}+1\right) x+a$ are

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

Sum of the zeros $=\alpha+\beta$

$=\frac{1}{a}+a$

$=\frac{1}{a}+\frac{a \times a}{1 \times a}$

$=\frac{1+a^{2}}{a}$

and, $=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$

$=\frac{-\left(a^{2}+1\right)}{a}$

$=\frac{a^{2}+1}{a}$

Product of the zeros $=\alpha \times \beta$

$=\frac{1}{a} \times a$

= 1

And, $=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$

$=\frac{a}{a}$

= 1

Therefore,

Product of the zeros $=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$

Hence, the relation-ship between the zeros and coefficient are verified.

(ix) $h(s)=2 s^{2}-(1+2 \sqrt{2}) s+\sqrt{2}$

$h(s)=2 s^{2}-s-2 \sqrt{2} s+\sqrt{2}$

$h(s)=s(2 s-1)-\sqrt{2}(2 s-1)$

h(s)=(2 s-1)(s-\sqrt{2})

The zeros of h(s) are given by 

h(s) = 0

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

$(2 s-1)(s-\sqrt{2})=0$

$(2 s-1)=0$ or $(s-\sqrt{2})=0$

$s=\frac{1}{2}$ or $s=\sqrt{2}$

Thus, the zeros of $h(s)=(2 s-1)(s-\sqrt{2})$ are $\alpha=\frac{1}{2}$ and $\beta=\sqrt{2}$.

Now,

Sum of the zeros $=\alpha+\beta$

$=\frac{1}{2}+\sqrt{2}$

and

$\frac{-\text { Coefficient of } s}{\text { Coefficient of } s^{2}}$

$=\frac{1+2 \sqrt{2}}{2}$

$=\frac{1}{2}+\sqrt{2}$

Therefore, sum of the zeros $=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$

Product of the zeros $=\alpha \beta$

$=\frac{1}{2} \times \sqrt{2}=\frac{1}{\sqrt{2}}$

and

$\frac{\text { Constant term }}{\text { Coefficient of } s^{2}}$

$=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$

Therefore,

Product of the zeros $=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$

Hence, the relation-ship between the zeros and coefficient are verified.

(x) $f(v)=v^{2}+4 \sqrt{3} v-15$

$f(v)=v^{2}+5 \sqrt{3} v-\sqrt{3} v-15$

$=v^{2}-\sqrt{3} v+5 \sqrt{3} v-15$

$=v(v-\sqrt{3})+5 \sqrt{3}(v-\sqrt{3})$

$=(v-\sqrt{3})(v+5 \sqrt{3})$

The zeros of f(v) are given by 

f(v) = 0

$v^{2}+4 \sqrt{3} v-15=0$

$(v+5 \sqrt{3})(v-\sqrt{3})=0$

$(v-\sqrt{3})=0$ or $(v+5 \sqrt{3})=0$

$v=\sqrt{3}$ or $v=-5 \sqrt{3}$

Thus, the zeros of $f(v)=(v-\sqrt{3})(v+5 \sqrt{3})$ are $\alpha=\sqrt{3}$ and $\beta=-5 \sqrt{3}$.

Now,

Sum of the zeros $=\alpha+\beta$

$=\sqrt{3}-5 \sqrt{3}=-4 \sqrt{3}$

and

$\frac{\text { - Coefficient of } v}{\text { Coefficient of } v^{2}}$

$=\frac{-4 \sqrt{3}}{1}=-4 \sqrt{3}$

Therefore, sum of the zeros $=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$

Product of the zeros $=\alpha \beta$

$=\sqrt{3} \times(-5 \sqrt{3})=-15$

and

$\frac{\text { Constant term }}{\text { Coefficient of } v^{2}}$

$=\frac{-15}{1}=-15$

Therefore,

Product of the zeros $=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$

Hence, the relation-ship between the zeros and coefficient are verified.

(xi) $p(y)=y^{2}+\frac{3 \sqrt{5}}{2} y-5$

$p(y)=\frac{1}{2}\left(2 y^{2}+4 \sqrt{5} y-\sqrt{5} y-10\right)$

$=\frac{1}{2}[2 y(y+2 \sqrt{5})-\sqrt{5}(y+2 \sqrt{5})]$

$=\frac{1}{2}[(2 y-\sqrt{5})(y+2 \sqrt{5})]$

The zeros are given by p(y) = 0.

Thus, the zeros of $p(y)=\frac{1}{2}(2 y-\sqrt{5})(y+2 \sqrt{5})$ are $\alpha=\frac{\sqrt{5}}{2}$ and $\beta=-2 \sqrt{5}$.

Now,

Sum of the zeros $=\alpha+\beta$

$=\frac{\sqrt{5}}{2}-2 \sqrt{5}=\frac{\sqrt{5}-4 \sqrt{5}}{2}=\frac{-3 \sqrt{5}}{2}$

and

$\frac{\text { - Coefficient of } y}{\text { Coefficient of } y^{2}}$

$=\frac{\frac{-3 \sqrt{5}}{2}}{1}=\frac{-3 \sqrt{5}}{2}$

Therefore, sum of the zeros $=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$

Product of the zeros $=\alpha \beta$

$=\frac{\sqrt{5}}{2} \times-2 \sqrt{5}$

$=-5$

and

$\frac{\text { Constant term }}{\text { Coefficient of } y^{2}}$

$=\frac{-5}{1}=-5$

Therefore,

Product of the zeros $=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$

Hence, the relation-ship between the zeros and coefficient are verified.

(xii) $q(y)=7 y^{2}-\frac{11}{3} y-\frac{2}{3}$

$q(y)=\frac{1}{3}\left(21 y^{2}-11 y-2\right)$

$=\frac{1}{3}\left(21 y^{2}-14 y+3 y-2\right)$

$=\frac{1}{3}[7 y(3 y-2)+1(3 y-2)]$

$=\frac{1}{3}[(7 y+1)(3 y-2)]$

The zeros are given by q(y) = 0.

Thus, the zeros of $q(y)=\frac{1}{3}(7 y+1)(3 y-2)$ are $\alpha=\frac{-1}{7}$ and $\beta=\frac{2}{3}$.

Now,

Thus, the zeros of $q(y)=\frac{1}{3}(7 y+1)(3 y-2)$ are $\alpha=\frac{-1}{7}$ and $\beta=\frac{2}{3}$.

Now,

Sum of the zeros $=\alpha+\beta$

$=\frac{-1}{7}+\frac{2}{3}$

$=\frac{11}{21}$

and

$\frac{-\text { Coefficient of } y}{\text { Coefficient of } y^{2}}$

$=\frac{-\left(-\frac{11}{3}\right)}{7}=\frac{11}{21}$

Therefore, sum of the zeros $=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$

Product of the zeros $=\alpha \beta$

$=\frac{-1}{7} \times \frac{2}{3}$

$=\frac{-2}{21}$

and

$\frac{\text { Constant term }}{\text { Coefficient of } y^{2}}$

$=\frac{\frac{-2}{3}}{7}$

$=\frac{-2}{21}$

Therefore,

Product of the zeros $=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$

 

 

 

 

ht = t2 - 15ht = t2 - 152ht = t + 15 t - 15">h(t) = t2 − 15