Chapter 10, Problem 7CA

To determine

To find the number of zeros.

Answer to Problem 7CA

  $\begin{array}{ccc}Nozeros& Onezero& Twozeros\\ \begin{array}{l}f\left(x\right)=3x^2+4x+2\\ f\left(x\right)=-6x^2-5\end{array}& \begin{array}{l}f\left(x\right)=4x^2-8x+4\\ f\left(x\right)=7x^2\end{array}& \begin{array}{l}f\left(x\right)=-x^2+2x\\ f\left(x\right)=x^2-3x-21\end{array}\end{array}$

Explanation of Solution

Given information: $f\left(x\right)=3x^2+4x+2$ , $f\left(x\right)=4x^2-8x+4$ , $f\left(x\right)=7x^2$ , $f\left(x\right)=-x^2+2x$ , $f\left(x\right)=x^2-3x-21$ and $f\left(x\right)=-6x^2-5$

Formula Used: $D=b^2-4ac$

Calculation:

Discriminant be used to confirm the number of real solutions of a quadratic equation because discriminant method is much easier and shorter than completing the square method.

We can easily calculate discriminant by $D=b^2-4ac$ and can know how many real numbers equation has.

• If $D$ equal to negative then $x$ has no real solution.
• If $D$ equal to 0 then $x$ has 1 real solution.
• If $D$ equal to positive then $x$ has 2 real solution.

(1)

  $f\left(x\right)=3x^2+4x+2$

As, $D=b^2-4ac$

Now, $a=3$ , $b=4$ and $c=2$

  $D=4^2-4\times 3\times 2$

So,

  $D=16-24$

Hence,

  $D=-8$

  $D$ is equal to negative then $x$ has no real solution.

(2)

  $f\left(x\right)=4x^2-8x+4$

As, $D=b^2-4ac$

Now, $a=4$ , $b=-8$ and $c=4$

  $D={\left(-8\right)}^2-4\times 4\times 4$

So,

  $D=64-64$

Hence,

  $D=0$

  $D$ is equal to 0 then $x$ has 1 real solution.

(3)

  $f\left(x\right)=7x^2$

As, $D=b^2-4ac$

Now, $a=7$ , $b=0$ and $c=0$

  $D={\left(0\right)}^2-4\times 7\times 0$

So,

  $D=0-0$

Hence,

  $D=0$

  $D$ is equal to 0 then $x$ has 1 real solution.

(4)

  $f\left(x\right)=-x^2+2x$

As, $D=b^2-4ac$

Now, $a=-1$ , $b=2$ and $c=0$

  $D={\left(2\right)}^2-4\times \left(-1\right)\times 0$

So,

  $D=4-0$

Hence,

  $D=4$

  $D$ is equal to positive then $x$ has 2 real solution.

(5)

  $f\left(x\right)=x^2-3x-21$

As, $D=b^2-4ac$

Now, $a=1$ , $b=-3$ and $c=-21$

  $D={\left(-3\right)}^2-4\times 1\times \left(-21\right)$

So,

  $D=9+84$

Hence,

  $D=93$

  $D$ is equal to positive then $x$ has 2 real solution.

(6)

  $f\left(x\right)=-6x^2-5$

As, $D=b^2-4ac$

Now, $a=-6$ , $b=0$ and $c=-5$

  $D={\left(0\right)}^2-4\times \left(-6\right)\times \left(-5\right)$

So,

  $D=0-120$

Hence,

  $D=-120$

  $D$ is equal to negative then $x$ has no real solution.

Hence Solution,

  $\begin{array}{ccc}Nozeros& Onezero& Twozeros\\ \begin{array}{l}f\left(x\right)=3x^2+4x+2\\ f\left(x\right)=-6x^2-5\end{array}& \begin{array}{l}f\left(x\right)=4x^2-8x+4\\ f\left(x\right)=7x^2\end{array}& \begin{array}{l}f\left(x\right)=-x^2+2x\\ f\left(x\right)=x^2-3x-21\end{array}\end{array}$