Chapter 1.5, Problem 123E

(a)

To determine

To find:the roots of the equation using middle term factorization.

Answer to Problem 123E

The solution of the given equation $\text{x=5}$ and $\text{x=4}$ .

Explanation of Solution

Given: ${\text{x}}^{\text{2}}\text{-9x=20=0}$ .

Concept used:

The First it can solve the given equation through middle term factorization by taking common factor and if it doesn’t get factorize then use discriminant method to solve the given quadratic equation where $\text{x=}\frac{\text{-b±}\sqrt{{\text{b}}^{\text{2}}\text{-4ac}}}{\text{2a}}$$\text{x=}\frac{\text{-b±}\sqrt{{\text{b}}^{\text{2}}\text{-4ac}}}{\text{2a}}$ .

Calculation:

  ${\text{x}}^{\text{2}}\text{-9x=20=0}$ .

First by solving the equation through middle term factorization

If it doesn’t get factorized then it can be solved from discriminant method to solve the given quadratic equation.

Where the equation is defined by ${\text{x}}^{\text{2}}\text{+sx+p=0}$

Where s =sum of the two roots and p= product of two roots

Let the two roots be $a,$$\text{b}$

Sum of the roots = $\text{s=a+b}$ ; product of the roots= $\text{p=a×b}$

Here in given solution $s=\left(\text{-5}\right)\text{+}\left(\text{-4}\right)\text{=-9}$ , $\text{p=}\left(\text{-4}\right)\text{×}\left(-5\right)\text{=20}$ .

Here need to find 2 number whose sum is $9$ and product is $20$ .

Which is possible by

  $\left(\text{-5}\right)\text{+}\left(\text{-4}\right)\text{=-9}$ .

  $\left(\text{-4}\right)\text{×}\left(-5\right)\text{=20}$ .

So, two numbers are $-5$ and $-4$ through this two numbers given quadratic equation can be factorized as:

  ${\text{x}}^{\text{2}}\text{-9x=20=0}$

  ${\text{x}}^{\text{2}}\text{-5x-4x-20=0}$

Since, $\left(\text{-5}\right)\text{+}\left(\text{-4}\right)\text{=-9}$ and $\left(\text{-4}\right)\text{×}\left(-5\right)\text{=20}$ .

  ${\text{x}}^{\text{2}}\text{-5x-4x-20=0}$ .

  $\text{x(x-5)-4(x-5)=0}$

  $\text{(x-5)(x-4)=0}$

So, $x=5,4$

Hence, the solution of the given equation $\text{x=5}$ and $\text{x=4}$ .

(b)

To determine

To find:the roots of the equation using middle term factorization or by quadratic formula.

Answer to Problem 123E

The solution of $\text{x=4,-2}$ .

Explanation of Solution

Given: ${\text{x}}^{\text{2}}\text{-2x-8=0}$ .

Concept used:

The First it can solve the given equation through middle term factorization by taking common factor and if it doesn’t get factorize then use discriminant method to solve the given quadratic equation where $\text{x=}\frac{\text{-b±}\sqrt{{\text{b}}^{\text{2}}\text{-4ac}}}{\text{2a}}$ .

Calculation:

  ${\text{x}}^{\text{2}}\text{-2x-8=0}$ .

Here the equation is defined by ${\text{x}}^{\text{2}}\text{+sx+p=0}$

Where s =sum of the two roots and p= product of two roots

Let the two roots be $a,$$\text{b}$

Sum of the roots = $\text{s=a+b}$ ; product of the roots= $\text{p=a×b}$

Here in given solution $s=\left(\text{-2}\right)\text{+}\left(\text{4}\right)\text{=2}$ , $\text{p=}\left(\text{-2}\right)\text{×}\left(4\right)\text{=8}$ .

This is not possible since sum of the root and product of the root is not equal so it cannot be solved from middle term factorization method.

It can be solved from discriminant method:

  $\text{x=}\frac{\text{-b±}\sqrt{{\text{b}}^{\text{2}}\text{-4ac}}}{\text{2a}}$ .

  ${\text{x}}^{\text{2}}\text{-2x-8=0}$ .

Here $\text{a=1, b=-2, c=-8}$ .

  $\text{x=}\frac{\text{-b±}\sqrt{{\text{b}}^{\text{2}}\text{-4ac}}}{\text{2a}}\text{=}\frac{\text{2±}\sqrt{\text{4+32}}}{\text{2}}\text{=}\frac{\text{2±6}}{\text{2}}\text{=1±3}$ .

  $\text{x=4,-2}$ .

Hence the solution of $\text{x=4,-2}$ .

(c)

To determine

To prove:that the quadratic equation has two roots which when multiply give product of the root C and which when sum give sum of the root B.

Answer to Problem 123E

  ${\text{c=r}}_{\text{1}}{\text{r}}_{\text{2}}$ and $\text{b=-}\left({\text{r}}_{\text{1}}{\text{+r}}_{\text{2}}\right)$ .

It is true that two roots which when multiply give product of the root C and which when sum give sum of the root B.

Explanation of Solution

Given: ${\text{x}}^{\text{2}}\text{+bx+c=0}$ .

  ${\text{c=r}}_{\text{1}}{\text{r}}_{\text{2}}$ and $\text{b=-}\left({\text{r}}_{\text{1}}{\text{+r}}_{\text{2}}\right)$ .

Concept used:

Using discriminant method to solve the given quadratic equation

where $\text{x=}\frac{\text{-b±}\sqrt{{\text{b}}^{\text{2}}\text{-4ac}}}{\text{2a}}$ .

Calculation:

  ${\text{x}}^{\text{2}}\text{+bx+c=0}$ .

  $\text{x=}\frac{\text{-b±}\sqrt{{\text{b}}^{\text{2}}\text{-4ac}}}{\text{2a}}$ .

  $\left(\frac{\text{-b+}\sqrt{{\text{b}}^{\text{2}}\text{-4ac}}}{\text{2}}\right)\text{.}\left(\frac{\text{-b-}\sqrt{{\text{b}}^{\text{2}}\text{-4ac}}}{\text{2}}\right)\text{=}\frac{{\text{b}}^{\text{2}}\text{-}\left({\text{b}}^{\text{2}}\text{-4c}\right)}{\text{4}}\text{=c}$ .

  $\text{-}\left({\text{r}}_{\text{1}}{\text{+r}}_{\text{2}}\right)\text{=-}\left(\frac{\text{-b+}\sqrt{{\text{b}}^{\text{2}}\text{-4ac}}}{\text{2}}\text{+}\frac{\text{-b-}\sqrt{{\text{b}}^{\text{2}}\text{-4ac}}}{\text{2}}\right)\text{=b}$ .