Chapter 9.7, Problem 40PFA

(A)

To determine

Solve the equation.

Answer to Problem 40PFA

No

Explanation of Solution

Given Information: $y=x^2-3x+1$ and $y=x+2$

Formula Used: $D=b^2-4ac$ and $x=\frac{-b\pm \sqrt{D}}{2a}$

Calculation:

• $D$ equal to negative so $x$ has no real roots.
• $D$ equal to positive so $x$ has real roots.

  $y=x^2-3x+1$ and $y=x+2$

Now, equate

  $x^2-3x+1=x+2$

Now,

  $x^2-4x-1=0$

Now, use formula $D=b^2-4ac$ to check if root of $x$ exist or not.

As $a=1,b=-4$ and $c=-1$

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

Now,

  $D=16+4$

Hence,

  $D=20$

As, $D$ equal to positive so $x$ has real roots.

  $x=\frac{-b\pm \sqrt{D}}{2a}$

So,

  $x=\frac{-\left(-4\right)\pm \sqrt{20}}{2\times 1}$

So, simplify

  $x=\frac{4\pm 2\sqrt{5}}{2}$

Hence

  $x=2\pm \sqrt{5}$

Now,

  $x=2+\sqrt{5}$ or $x=2-\sqrt{5}$

(B)

To determine

Solve the equation.

Answer to Problem 40PFA

No

  $\left(x,y\right)=\left(-1,-3\right)$

Explanation of Solution

Given Information: $y=x^2+3x-1$ and $y=x-2$

Formula Used: $a^2+2ab+b^2={\left(a+b\right)}^2$

Calculation:

• $D$ equal to negative so $x$ has no real roots.
• $D$ equal to positive so $x$ has real roots.

  $y=x^2+3x-1$ and $y=x-2$

Now, equate

  $x^2+3x-1=x-2$

Now,

  $x^2+2x+1=0$

Now, Use formula $a^2+2ab+b^2={\left(a+b\right)}^2$

  ${\left(x+1\right)}^2=0$

Hence

  $\left(x+1\right)\left(x+1\right)=0$

Now,

  $x=-1$

So, find value of $y$

  $y=x-2$

Now, put $x=-1$

  $y=-1-2$

Hence,

  $y=-3$

So, $\left(x,y\right)=\left(-1,-3\right)$

(C)

To determine

Solve the equation.

Answer to Problem 40PFA

Yes

  $\left(x,y\right)=\left(0,-2\right),\left(1,-1\right)$

Explanation of Solution

Given Information: $y=x^2-2$ and $y=x-2$

Formula Used: $a^2+2ab+b^2={\left(a+b\right)}^2$

Calculation:

• $D$ equal to negative so $x$ has no real roots.
• $D$ equal to positive so $x$ has real roots.

  $y=x^2-2$ and $y=x-2$

Now, equate

  $x^2-2=x-2$

Now,

  $x^2-x=0$

Now, simplify

  $x\left(x-1\right)=0$

Hence

  $x=0$ or $x=1$

So, find value of $y$

  $y=x-2$

Now, put $x=0$ or $x=1$

  $y=0-2$ or $y=1-2$

Hence,

  $y=-2$ or $y=-1$

So, $\left(x,y\right)=\left(0,-2\right),\left(1,-1\right)$

(D)

To determine

Solve the equation.

Answer to Problem 40PFA

No

Explanation of Solution

Given Information: $y=x^2-2$ and $y=-x+2$

Formula Used: $D=b^2-4ac$ and $x=\frac{-b\pm \sqrt{D}}{2a}$

Calculation:

• $D$ equal to negative so $x$ has no real roots.
>
• $D$ equal to positive so $x$ has real roots.

  $y=x^2-2$ and $y=-x+2$

Now, equate

  $x^2-2=-x+2$

Now,

  $x^2+x-4=0$

Now, use formula $D=b^2-4ac$ to check if root of $x$ exist or not.

As $a=1,b=1$ and $c=-4$

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

Now,

  $D=1+16$

Hence,

  $D=17$

As, $D$ equal to positive so $x$ has real roots.

  $x=\frac{-b\pm \sqrt{D}}{2a}$

So,

  $x=\frac{-1\pm \sqrt{17}}{2\times 1}$

So, simplify

  $x=\frac{-1\pm \sqrt{17}}{2}$