Chapter 9.4, Problem 66E

a.

To determine

To estimate the value of $x$ using graphing method.

Answer to Problem 66E

  $x=3$ and $x=1$

Explanation of Solution

Given:

Equation:

  $y=x^2-4x+6$

  $y=3$

Calculation:

For $y=3$ find $x$

Substituting the value of y ,

  $3=x^2-4x+6$

Changing the equation in standard form,

  $x^2-4x+6-3=0$

So,

  $x^2-4x+3=0$

Splitting the middle term in such a way that the sum comes -4 and product comes 3

  $x(x-1)-3(x-1)=0$

So,

  $(x-3)(x-1)=0$

So,

  $(x-3)=0$ and $(x-1)=0$

So,

  $x=3$ and $x=1$

b.

To determine

To use completing the square method to check the estimating part.

Answer to Problem 66E

The estimation was correct.

Explanation of Solution

Given:

Equation:

  $y=x^2-4x+6$

Calculation:

Changing the equation in standard form,

  $x^2-4x+6-3=0$

So,

  $x^2-4x+3=0$

Taking constant on one side

  $x^2-4x=-3$

Taking square half of coefficient of $x$

i.e. ${(4\times \frac{1}{2})}^2$

so,

  ${(2)}^2=4$

Adding this on both side,

  $x^2-4x+4=-3+4$

  $x^2-4x+4$ can be written as ${(x-2)}^2$

Therefore,

  ${(x-2)}^2=-3+4$

Solving right hand side,

  ${(x-2)}^2=1$

So,

  $x-2=\sqrt{1}$

Solving root,

  $x-2=\pm 1$

Therefore ,

  $x=2\pm 1$

So,

  $x=2+1$ and $x=2-1$

So,

  $x=3$ and $x=1$

Now put $x=3$ in equation $y=x^2-4x+6$

  $y={(3)}^2-4(3)+6$

So,

  $y=9-12+6$

So,

  $y=3$

Again, put $x=1$ in equation $y=x^2-4x+6$

  $y={(1)}^2-4(1)+6$

So,

  $y=1-4+6$

So,

  $y=3$

So the estimation was correct.