Chapter 9.4, Problem 47PPS

(a)

To determine

Copy the table shown and complete the second column.

Answer to Problem 47PPS

Trinomial	$b^2-4ac$	Number of roots
$x^2-8x+16$	${(-8)}^2-4·1·16=0$	1
$2x^2-11x+3$	${(-11)}^2-4·2·3=97$	2
$3x^2+6x+9$	${(6)}^2-4·3·9=-72$	0
$x^2-2x+7$	${(-2)}^2-4·1·7=-24$	0
$x^2+10x+25$	${(10)}^2-4·1·25=0$	1
$x^2+3x-12$	${(3)}^2-4·3·-12=57$	2

Explanation of Solution

Given:

The trinomials are: $x^2-8x+16$ ; $2x^2-11x+3$ ; $3x^2+6x+9$ ; $x^2-2x+7$ ; $x^2+10x+25$ ; $x^2+3x-12$

Concept Used:

• The discriminant is given by, $D=b^2-4ac$
• For a quadratic equation $a{x}^2+bx+c=0$ the number of roots depend on:
• If discriminant is negative, $b^2-4ac<0$, then there is no real roots
• If discriminant is zero, $b^2-4ac=0$, then there are two same real roots
• If discriminant is positive, $b^2-4ac>0$, then there are two distinct real roots

Calculation:

Trinomial	$b^2-4ac$	Number of roots
$x^2-8x+16$	${(-8)}^2-4·1·16=0$	1
$2x^2-11x+3$	${(-11)}^2-4·2·3=97$	2
$3x^2+6x+9$	${(6)}^2-4·3·9=-72$	0
$x^2-2x+7$	${(-2)}^2-4·1·7=-24$	0
$x^2+10x+25$	${(10)}^2-4·1·25=0$	1
$x^2+3x-12$	${(3)}^2-4·3·-12=57$	2

(b)

To determine

Set each trinomial equal to zero, and solve the equation by completing the square. Complete the last column of the table with the number of roots of each equation.

Answer to Problem 47PPS

Trinomial	$b^2-4ac$	Number of roots
$x^2-8x+16$	${(-8)}^2-4·1·16=0$	1
$2x^2-11x+3$	${(-11)}^2-4·2·3=97$	2
$3x^2+6x+9$	${(6)}^2-4·3·9=-72$	0
$x^2-2x+7$	${(-2)}^2-4·1·7=-24$	0
$x^2+10x+25$	${(10)}^2-4·1·25=0$	1
$x^2+3x-12$	${(3)}^2-4·3·-12=57$	2

Explanation of Solution

Given:

The trinomials are: $x^2-8x+16$ ; $2x^2-11x+3$ ; $3x^2+6x+9$ ; $x^2-2x+7$ ; $x^2+10x+25$ ; $x^2+3x-12$

Calculation:

Solve the equation by completing the square.

  $\begin{array}{l}{x}^2-8x+16=0[\text{original}\text{equation}]\\ x^2-8x=-16[\text{Subtract}16\text{from}\text{each}\text{side}]\\ $ x^2-8x+16=-16+16[\text{Since}{(\frac{8}{2})}^2=16\text{add}\text{to}\text{each}\text{side}]$$ {(x-4)}^2=0[\text{Factor}]$\sqrt{{(x-4)}^2}=\pm \sqrt{0}[\text{Take}\text{the}\text{square}\text{root}\text{of}\text{each}\text{side}]\\ x-4=0\\ x-4+4=4[\text{add}\text{4}\text{to}\text{both}\text{sides}]\\ x=4\end{array}$

The trinomial $x^2-8x+16$has 1 root.

  $\begin{array}{l}2x^2-11x+3=0[\text{original}\text{equation}]\\ 2x^2-11x=-3[\text{Subtract}3\text{from}\text{each}\text{side}]\\ \frac{2x^2-11x}{2}=-\frac{3}{2}[\text{Divide}\text{each}\text{side}\text{by}\text{2}]\end{array}$

  $x^2-\frac{11}{2}x+\frac{121}{4}=-\frac{3}{2}+\frac{121}{4}[\text{Since}{(\frac{11}{2})}^2=\frac{121}{4}\text{add}\text{to}\text{each}\text{side}]$

  ${(x-\frac{11}{4})}^2=\frac{115}{4}[\text{Factor}]$

  $\begin{array}{l}\sqrt{{(x-\frac{11}{4})}^2}=\pm \sqrt{\frac{115}{4}}[\text{Take}\text{the}\text{square}\text{root}\text{of}\text{each}\text{side}]\\ x-\frac{11}{4}=\pm \frac{\sqrt{115}}{2}\\ x-\frac{11}{4}+\frac{11}{4}=\frac{11}{4}\pm \frac{\sqrt{115}}{2}[\text{add}\frac{11}{4}\text{to}\text{both}\text{sides}]\\ x=\frac{11}{4}+\frac{\sqrt{115}}{2}\text{or}\frac{11}{4}-\frac{\sqrt{115}}{2}\\ x=8.11orx\approx -2.61\end{array}$

The trinomial $2x^2-11x+3$has 2roots.

  $\begin{array}{l}3x^2+6x+9=0[\text{original}\text{equation}]\\ 3x^2+6x=-9[\text{Subtract}9\text{from}\text{each}\text{side}]\\ \frac{3x^2+6x}{3}=-\frac{9}{3}[\text{Divide}\text{each}\text{side}\text{by}\text{3}]\\ $ x^2+2x+1=-3+1[\text{Since}{(\frac{2}{2})}^2=1\text{add}\text{to}\text{each}\text{side}]$$ {(x+1)}^2=-2[\text{Factor}]$\sqrt{{(x+1)}^2}=\pm \sqrt{-2}[\text{Take}\text{the}\text{square}\text{root}\text{of}\text{each}\text{side}]\\ x+1=\pm \sqrt{-2}\end{array}$

The square root of a negative number has no real roots. So, the trinomial $3x^2+6x+9$has 0 roots.

  $\begin{array}{l}{x}^2-2x+7=0[\text{original}\text{equation}]\\ x^2-2x=-7[\text{Subtract}7\text{from}\text{each}\text{side}]\\ $ x^2-2x+1=-7+1[\text{Since}{(\frac{2}{2})}^2=1\text{add}\text{to}\text{each}\text{side}]$$ {(x-1)}^2=-6[\text{Factor}]$\sqrt{{(x-1)}^2}=\pm \sqrt{-6}[\text{Take}\text{the}\text{square}\text{root}\text{of}\text{each}\text{side}]\\ x-1=\pm \sqrt{-6}\end{array}$

The square root of a negative number has no real roots. So, the trinomial $x^2-2x+7$has 0 roots.

  $\begin{array}{l}{x}^2+10x+25=0[\text{original}\text{equation}]\\ x^2+10x=-25[\text{Subtract}16\text{from}\text{each}\text{side}]\\ $ x^2+10x+16=-25+25[\text{Since}{(\frac{10}{2})}^2=25\text{add}\text{to}\text{each}\text{side}]$$ {(x+5)}^2=0[\text{Factor}]$\sqrt{{(x+5)}^2}=\pm \sqrt{0}[\text{Take}\text{the}\text{square}\text{root}\text{of}\text{each}\text{side}]\\ x+5=0\\ x+5-5=-5[\text{Subtract}\text{4}\text{to}\text{both}\text{sides}]\\ x=-5\end{array}$

The trinomial $x^2+10x+25$has 1 root.

  $\begin{array}{l}{x}^2+3x-12=0[\text{original}\text{equation}]\\ x^2+3x=12[\text{Add}12\text{from}\text{each}\text{side}]\end{array}$

  $x^2+3x+\frac{9}{4}=12+\frac{9}{4}[\text{Since}{(\frac{3}{2})}^2=\frac{9}{4}\text{add}\text{to}\text{each}\text{side}]$

  ${(x+\frac{3}{2})}^2=\frac{57}{4}[\text{Factor}]$

  $\begin{array}{l}\sqrt{{(x+\frac{3}{2})}^2}=\pm \sqrt{\frac{57}{4}}[\text{Take}\text{the}\text{square}\text{root}\text{of}\text{each}\text{side}]\\ x+\frac{3}{2}=\pm \frac{\sqrt{57}}{2}\\ x+\frac{3}{2}-\frac{3}{2}=-\frac{3}{2}\pm \frac{\sqrt{57}}{2}[\text{subtract}\frac{3}{2}\text{to}\text{both}\text{sides}]\\ x=-\frac{3}{2}+\frac{\sqrt{57}}{2}\text{or}-\frac{3}{2}-\frac{\sqrt{57}}{2}\\ x\approx 2.3\text{or}x\approx -5.3\end{array}$

The trinomial $x^2+3x-12$has 2 roots.

Trinomial	$b^2-4ac$	Number of roots
$x^2-8x+16$	${(-8)}^2-4·1·16=0$	1
$2x^2-11x+3$	${(-11)}^2-4·2·3=97$	2
$3x^2+6x+9$	${(6)}^2-4·3·9=-72$	0
$x^2-2x+7$	${(-2)}^2-4·1·7=-24$	0
$x^2+10x+25$	${(10)}^2-4·1·25=0$	1
$x^2+3x-12$	${(3)}^2-4·3·-12=57$	2

Thus,

(c)

To determine

Compare the number of roots of each equation to the result in the $b^2-4ac$ column.

Answer to Problem 47PPS

If $b^2-4ac$ is negative, the equation has no real solutions. If $b^2-4ac$ is zero, the equation has one solution. If $b^2-4ac$ is positive, the equation has 2 solutions.

Explanation of Solution

Given:

The trinomials are: $x^2-8x+16$ ; $2x^2-11x+3$ ; $3x^2+6x+9$ ; $x^2-2x+7$ ; $x^2+10x+25$ ; $x^2+3x-12$

Concept Used:

If $b^2-4ac$ is negative, the equation has no real solutions. If $b^2-4ac$ is zero, the equation has one solution. If $b^2-4ac$ is positive, the equation has 2 solutions.

Thus, if $b^2-4ac$ is negative, the equation has no real solutions. If $b^2-4ac$ is zero, the equation has one solution. If $b^2-4ac$ is positive, the equation has 2 solutions.

(d)

To determine

Predict how many solutions $2x^2-9x+15=0$ will have. Verify by solving the equation.

Answer to Problem 47PPS

The equation $2x^2-9x+15=0$ has 0 real solution because $b^2-4ac$ is negative

Explanation of Solution

Given:

The trinomials are: $x^2-8x+16$ ; $2x^2-11x+3$ ; $3x^2+6x+9$ ; $x^2-2x+7$ ; $x^2+10x+25$ ; $x^2+3x-12$

Concept Used:

  $\begin{array}{l}\text{Equation}:2x^2-9x+15=0\\ $ b^2-4ac={(-9)}^2-4·2·15=81-120=-39$\end{array}$

The equation $2x^2-9x+15=0$ has 0 real solution because $b^2-4ac$ is negative

Thus, the equation $2x^2-9x+15=0$ has 0 real solution because $b^2-4ac$ is negative