Question

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

(i) f(x) = x² − 2x − 8
(ii) g(s) = 4s² − 4s + 1
(iii) h(t) = t² − 15
(iv) 6x² − 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)$f\left(x\right)=x^2-\left(\sqrt{3}+1\right)x+\sqrt{3}$
(viii) g(x) = a(x² + 1) − x(a² + 1)
(ix) $h\left(s\right)=2s^2-\left(1+2\sqrt{2}\right)s+\sqrt{2}$
(x) $f\left(v\right)=v^2+4\sqrt{3}v-15$
(xi) $p\left(y\right)=y^2+\frac{3\sqrt{5}}{2}y-5$
(xii) $q\left(y\right)=7y^2-\frac{11}{3}y-\frac{2}{3}$

Answer 1

(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,

and

Therefore, sum of the zeros =

Product of the zeros

_{= − 2 × 4}

_{= −8}

and

Therefore,

Product of the zeros =

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

(ii) Given

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

Or

Thus, the zeros ofare

and

Now, sum of the zeros

and

Therefore, sum of the zeros =

Product of the zeros

and =

Therefore, the product of the zeros =

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

(iii) Given

We have,

$$\begin{gathered} h\left(t\right)=t^2-15 \\ h\left(t\right)={\left(t\right)}^2-{\left(\sqrt{15}\right)}^2 \\ h\left(t\right)=\left(t+\sqrt{15}\right)\left(t-\sqrt{15}\right) \end{gathered}$$

The zeros of are given by

$$\begin{gathered} h\left(t\right)=0 \\ \left(t-\sqrt{15}\right)\left(t+\sqrt{15}\right)=0 \\ \left(t-\sqrt{15}\right)=0 \\ t=\sqrt{15} \\ \mathrm{or} \\ \left(t+\sqrt{15}\right)=0 \\ t=-\sqrt{15} \end{gathered}$$

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

Now,

Sum of the zeros

and =

Therefore, sum of the zeros =

also,

Product of the zeros = αβ

and,

Therefore, the product of the zeros =

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

(iv) Given

We have,

The zeros of are given by

Or

Thus, the zeros of are and.

Now,

Sum of the zeros = α + β

and, =

Therefore, sum of the zeros =

Product of the zeros = α × β

and, =

Product of zeros =

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

(v) Given

We have,

The zeros of are given by

Or

Thus, The zeros of areand

Now,

Sum of the zeros = α + β

and,

Therefore, Sum of the zeros =

Product of the zeros

and

Therefore, The product of the zeros =

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

(vi) Given

We have,

The zeros of g(x) are given by

Or

Thus, the zeros of are and.

Now,

Sum of the zeros = α + β

and =

Therefore, sum of the zeros =

Product of zeros = α × β

and =

Therefore, the product of the zeros =

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

(vii) Given

The zeros of ƒ(x) are given by

Or

Thus, the zeros of are α = 1 and

Now,

Sum of zeros = α + β

And,

Therefore, sum of the zeros =

Product of the zeros = αβ

And

=

Product of zeros =

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

(viii) Given

The zeros of g(x) are given by

or

Thus, the zeros of are

and

Sum of the zeros = α + β

and, =

Product of the zeros

And, =

Therefore,

Product of the zeros =

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

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

$$\begin{gathered} h\left(s\right)=2s^2-s-2\sqrt{2}s+\sqrt{2} \\ h\left(s\right)=s\left(2s-1\right)-\sqrt{2}\left(2s-1\right) \\ h\left(s\right)=\left(2s-1\right)\left(s-\sqrt{2}\right) \end{gathered}$$

The zeros of h(s) are given by

h(s) = 0

$$\begin{gathered} 2s^2-\left(1+2\sqrt{2}\right)s+\sqrt{2}=0 \\ \left(2s-1\right)\left(s-\sqrt{2}\right)=0 \\ \left(2s-1\right)=0\mathrm{or}\left(s-\sqrt{2}\right)=0 \\ s=\frac{1}{2}\mathrm{or}s=\sqrt{2} \end{gathered}$$

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

Now,

Sum of the zeros = $\alpha +\beta$
$=\frac{1}{2}+\sqrt{2}$

and

$$\begin{gathered} \frac{-\mathrm{Coefficient}\mathrm{of}s}{\mathrm{Coefficient}\mathrm{of}{s}^2} \\ =-\left(\frac{-\left(1+2\sqrt{2}\right)}{2}\right) \\ =\frac{1+2\sqrt{2}}{2} \\ =\frac{1}{2}+\sqrt{2} \end{gathered}$$

Therefore, sum of the zeros =

Product of the zeros
$=\frac{1}{2}\times \sqrt{2}=\frac{1}{\sqrt{2}}$

and

$$\begin{gathered} \frac{\mathrm{Constant}\mathrm{term}}{\mathrm{Coefficient}\mathrm{of}{s}^2} \\ =\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}} \end{gathered}$$

Therefore,

Product of the zeros =

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

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

$$\begin{gathered} f\left(v\right)=v^2+5\sqrt{3}v-\sqrt{3}v-15 \\ =v^2-\sqrt{3}v+5\sqrt{3}v-15 \\ =v\left(v-\sqrt{3}\right)+5\sqrt{3}\left(v-\sqrt{3}\right) \\ =\left(v-\sqrt{3}\right)\left(v+5\sqrt{3}\right) \end{gathered}$$

The zeros of f(v) are given by

f(v) = 0

$$\begin{gathered} v^2+4\sqrt{3}v-15=0 \\ \left(v+5\sqrt{3}\right)\left(v-\sqrt{3}\right)=0 \\ \left(v-\sqrt{3}\right)=0\mathrm{or}\left(v+5\sqrt{3}\right)=0 \\ v=\sqrt{3}\mathrm{or}v=-5\sqrt{3} \end{gathered}$$

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

Now,

Sum of the zeros = $\alpha +\beta$
$=\sqrt{3}-5\sqrt{3}=-4\sqrt{3}$

and

$$\begin{gathered} \frac{-\mathrm{Coefficient}\mathrm{of}v}{\mathrm{Coefficient}\mathrm{of}{v}^2} \\ =\frac{-4\sqrt{3}}{1}=-4\sqrt{3} \end{gathered}$$

Therefore, sum of the zeros =

Product of the zeros
$=\sqrt{3}\times \left(-5\sqrt{3}\right)=-15$

and

$$\begin{gathered} \frac{\mathrm{Constant}\mathrm{term}}{\mathrm{Coefficient}\mathrm{of}{v}^2} \\ =\frac{-15}{1}=-15 \end{gathered}$$

Therefore,

Product of the zeros =

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

(xi) $p\left(y\right)=y^2+\frac{3\sqrt{5}}{2}y-5$
$$\begin{gathered} p\left(y\right)=\frac{1}{2}\left(2y^2+4\sqrt{5}y-\sqrt{5}y-10\right) \\ =\frac{1}{2}\left[2y\left(y+2\sqrt{5}\right)-\sqrt{5}\left(y+2\sqrt{5}\right)\right] \\ =\frac{1}{2}\left[\left(2y-\sqrt{5}\right)\left(y+2\sqrt{5}\right)\right] \end{gathered}$$
The zeros are given by p(y) = 0.
Thus, the zeros of $p\left(y\right)=\frac{1}{2}\left(2y-\sqrt{5}\right)\left(y+2\sqrt{5}\right)\mathrm{are}\alpha =\frac{\sqrt{5}}{2}\mathrm{and}\beta =-2\sqrt{5}$.

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

$$\begin{gathered} \frac{-\mathrm{Coefficient}\mathrm{of}y}{\mathrm{Coefficient}\mathrm{of}{y}^2} \\ =\frac{{\displaystyle \frac{-3\sqrt{5}}{2}}}{1}=\frac{-3\sqrt{5}}{2} \end{gathered}$$

Therefore, sum of the zeros =

Product of the zeros
$$\begin{gathered} =\frac{\sqrt{5}}{2}\times -2\sqrt{5} \\ =-5 \end{gathered}$$

and

$$\begin{gathered} \frac{\mathrm{Constant}\mathrm{term}}{\mathrm{Coefficient}\mathrm{of}{y}^2} \\ =\frac{-5}{1}=-5 \end{gathered}$$

Therefore,

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

(xii) $q\left(y\right)=7y^2-\frac{11}{3}y-\frac{2}{3}$
$$\begin{gathered} q\left(y\right)=\frac{1}{3}\left(21y^2-11y-2\right) \\ =\frac{1}{3}\left(21y^2-14y+3y-2\right) \\ =\frac{1}{3}\left[7y\left(3y-2\right)+1\left(3y-2\right)\right] \\ =\frac{1}{3}\left[\left(7y+1\right)\left(3y-2\right)\right] \end{gathered}$$

The zeros are given by q(y) = 0.

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

Now,

Sum of the zeros = $\alpha +\beta$
$$\begin{gathered} =\frac{-1}{7}+\frac{2}{3} \\ =\frac{11}{21} \end{gathered}$$

and

$$\begin{gathered} \frac{-\mathrm{Coefficient}\mathrm{of}y}{\mathrm{Coefficient}\mathrm{of}{y}^2} \\ =\frac{-\left(-{\displaystyle \frac{11}{3}}\right)}{7}=\frac{11}{21} \end{gathered}$$

Therefore, sum of the zeros =

Product of the zeros
$$\begin{gathered} =\frac{-1}{7}\times \frac{2}{3} \\ =\frac{-2}{21} \end{gathered}$$

and

$$\begin{gathered} \frac{\mathrm{Constant}\mathrm{term}}{\mathrm{Coefficient}\mathrm{of}{y}^2} \\ =\frac{{\displaystyle \frac{-2}{3}}}{7} \\ =\frac{-2}{21} \end{gathered}$$

Therefore,

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