Chapter A4, Problem 1CP

(a)

To determine

To calculate: The real zeros of polynomial, $\left(2x^2+4x+1\right)$.

Answer to Problem 1CP

Solution:

The real zeros for polynomial $\left(2x^2+4x+1\right)$ are $\frac{-2+\sqrt{2}}{2}\text{ and }\frac{-2-\sqrt{2}}{2}$.

Explanation of Solution

Given information:

The provided polynomial is $\left(2x^2+4x+1\right)$.

Formula used:

If $a{x}^2+bx+c=0$, where $a\ne 0$ is a quadratic equation, then real zeros (x) is given by,

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

Consider the polynomial,

$\left(2x^2+4x+1\right)$

Compare the quadratic equation $\left(2x^2+4x+1\right)$ with general quadratic equation $\left(a{x}^2+bx+c\right)$ and find out the value of a, b and c that is,

$a=2,b=4\text{ and }c=1$.

Now apply, the formula of real zeros for quadratic equation $a{x}^2+bx+c=0$,

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute, $2$ for $a$, $4$ for $b$ and $1$ for $c$ in above equation.

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

The real zeros for quadratic equation $\left(2x^2+4x+1\right)$,

$\begin{array}{c}x=\frac{-\left(4\right)\pm \sqrt{{\left(4\right)}^2-4\left(2\right)\left(1\right)}}{2\left(2\right)}\\ =\frac{-4\pm \sqrt{16-8}}{4}\\ =\frac{-4\pm \sqrt{8}}{4}\\ =\frac{-2\pm \sqrt{2}}{2}\end{array}$

Hence, the real zeros for polynomial $\left(2x^2+4x+1\right)$ are $\frac{-2+\sqrt{2}}{2}\text{ and }\frac{-2-\sqrt{2}}{2}$.

(b)

To determine

To calculate: The real zeros of polynomial, $\left(x^2-8x+16\right)$.

Answer to Problem 1CP

Solution:

The real zeros for polynomial $\left(x^2-8x+16\right)$ are $\text{4 and }4$.

Explanation of Solution

Given information:

The polynomial is $\left(x^2-8x+16\right)$.

Formula used:

If $a{x}^2+bx+c=0$, where $a\ne 0$ is a quadratic equation, then real zeros (x) is given by,

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

Consider the polynomial,

$\left(x^2-8x+16\right)$

Compare the quadratic equation $\left(x^2-8x+16\right)$ with general quadratic equation $\left(a{x}^2+bx+c\right)$ and find out the value of a, b and c that is,

$a=1,b=-8\text{ and }c=16$.

Now apply, the formula of real zeros for quadratic equation $a{x}^2+bx+c=0$,

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute, $1$ for $a$, $-8$ for $b$ and $16$ for $c$ in above equation.

$x=\frac{-\left(-8\right)\pm \sqrt{{\left(-8\right)}^2-4\left(1\right)\left(16\right)}}{2\left(1\right)}$

The real zeros for quadratic equation $\left(x^2-8x+16\right)$,

$\begin{array}{c}x=\frac{-\left(-8\right)\pm \sqrt{{\left(-8\right)}^2-4\left(1\right)\left(16\right)}}{2\left(1\right)}\\ =\frac{8\pm \sqrt{64-64}}{4}\\ =\frac{8}{2}\\ =4\end{array}$

Hence, the real zeros for polynomial $\left(x^2-8x+16\right)$ are $\text{4 and }4$.

(c)

To determine

To calculate: The real zeros of polynomial, $\left(2x^2-x+5\right)$.

Answer to Problem 1CP

Solution:

There are no real zeros for polynomial $\left(2x^2-x+5\right)$.

Explanation of Solution

Given information:

The polynomial is $\left(2x^2-x+5\right)$.

Formula used:

If $a{x}^2+bx+c=0$, where $a\ne 0$ is a quadratic equation, then real zeros (x) is given by,

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

The value of $i$ is $\sqrt{-1}$.

Here, complex number is $i$.

Calculation:

Consider the polynomial,

$\left(2x^2-x+5\right)$

Compare the quadratic equation $\left(2x^2-x+5\right)$ with general quadratic equation $\left(a{x}^2+bx+c\right)$ and find out the value of a, b and c that is,

$a=2,b=-1\text{ and }c=5$.

Now apply, the formula of real zeros for quadratic equation $a{x}^2+bx+c=0$,

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

The real zeros for quadratic equation $\left(2x^2-x+5\right)$ are given by

$x=\frac{-\left(-1\right)\pm \sqrt{{\left(-1\right)}^2-4\left(2\right)\left(5\right)}}{2\left(2\right)}$

The real zeros for quadratic equation $\left(2x^2-x+5\right)$,

$\begin{array}{c}x=\frac{-\left(-1\right)\pm \sqrt{{\left(-1\right)}^2-4\left(2\right)\left(5\right)}}{2\left(2\right)}\\ =\frac{1\pm \sqrt{1-40}}{4}\\ =\frac{1\pm \sqrt{-39}}{4}\\ =\frac{1\pm \sqrt{39}i}{4}\end{array}$

Since, the zeros of polynomial are complex number $\frac{1+\sqrt{39}i}{4}\text{ and }\frac{1-\sqrt{39}i}{4}$.

Thus, there is no real zeros for polynomial $\left(2x^2-x+5\right)$.