Quadratic Inequalities 2

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Question 1 of 4

1. Question
Solve for $x$ $x$

$3 x^{2} - 3 < 0$ $3x^2-3<0$
- 1. $x$ $x$ $<$ $<$ $- 3$ $-3$ and $x$ $x$ $>$ $>$ $3$ $3$
- 2. $- 1$ $-1$ $<$ $<$ $x$ $x$ $<$ $<$ $1$ $1$
- 3. $x$ $x$ $<$ $<$ $- 1$ $-1$ and $x$ $x$ $>$ $>$ $1$ $1$
- 4. $- 3$ $-3$ $<$ $<$ $x$ $x$ $<$ $<$ $3$ $3$
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The difference of two squares, $a^{2} - b^{2}$ $a^2-b^2$ , can be factored as the sum and and difference of $a$ $a$ and $b$ $b$ $(a + b) (a - b)$ $(a+b)(a-b)$

First, change the inequality sign into an equal sign and find the $x$ $x$ values

$3 x^{2} - 3$ $3x^2-3$ $=$ $=$ $0$ $0$

$3 (x^{2} - 1)$ $3(x^2-1)$ $=$ $=$ $0$ $0$ Factor out $3$ $3$

$3 (x + 1) (x - 1)$ $3(x+1)(x-1)$ $=$ $=$ $0$ $0$ Difference of two squares

$x + 1$ $x+1$ $=$ $=$ $0$ $0$

$x + 1$ $x+1$ $- 1$ $-1$ $=$ $=$ $0$ $0$ $- 1$ $-1$

$x$ $x$ $=$ $=$ $- 1$ $-1$

$x - 1$ $x-1$ $=$ $=$ $0$ $0$

$x - 1$ $x-1$ $+ 1$ $+1$ $=$ $=$ $0$ $0$ $+ 1$ $+1$

$x$ $x$ $=$ $=$ $1$ $1$

Mark these $2$ $2$ points on the $x$ $x$ axis.

Next, substitute $x = 0$ $x=0$ to the function to get the $y$ $y$ intercept

$y$ $y$ $=$ $=$ $3 x^{2} - 3$ $3x^2-3$

$y$ $y$ $=$ $=$ $3 {(0)}^{2} - 3$ $3(0)^2-3$ Substitute $x=0$

$y$ $=$ $0-3$

$y$ $=$ $-3$

Mark this point on the $y$ axis.

Form a parabola by connecting the points

Since we are looking for $y$ $<$ $0$ , the values are below the $x$ axis

Hence, $-1$ $<$ $x$ $<$ $1$

$-1$ $<$ $x$ $<$ $1$

Question 2 of 4

Solve for

$x$ :

$2x^2+3x-7≥0$

1. $x≤-2.766$ and $x≥1.266$
2. $-2.766≤x≤1.266$
3. $x≥1.266$
4. $x≤-2.766$

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The Quadratic Formula

$x=\frac {-\color{#9a00c7}{b} \pm \sqrt {\color{#9a00c7}{b}^2-4\color{#00880A}{a}\color{#007DDC}{c}} }{2 \color{#00880A}{a}}$

First, replace the inequality with an equal sign and solve for

$x$ using the Quadratic Formula

$2x^2+3x-7=0$

$a=2$ $b=3$ $c=-7$

$x$	$=$	$\frac {-\color{#9a00c7}{b} \pm \sqrt {\color{#9a00c7}{b}^2-4\color{#00880A}{a}\color{#007DDC}{c}} }{2 \color{#00880A}{a}}$	Quadratic Formula

	$=$	$\frac {- \color{#9a00c7}{3} \pm \sqrt {\color{#9a00c7}{3}^2-4\color{#00880A}{(2)}\color{#007DDC}{(-7)}} }{2 \color{#00880A}{(2)}}$	Plug in the values of $a, b$ and $c$

	$=$	$\frac {-3 \pm \sqrt {9 +56} }{4}$

	$=$	$\frac {-3 \pm \sqrt {65} }{4}$

Write each root individually

$x_1$	$=$	$\frac {-3 + \sqrt {65} }{4}$

	$=$	$1.266$

$x_2$	$=$	$\frac {-3 – \sqrt {65} }{4}$

	$=$	$-2.766$

Mark these two points on the

$x$ axis

Next, find the

$y$ intercept by substituting

$x=0$

$y$	$=$	$2x^2+3x-7$
$y$	$=$	$2(0)^2+3(0)-7$	Substitute $x=0$
$y$	$=$	$0-0-7$
$y$	$=$	$-7$

Mark this on the

$y$ axis

Form a parabola by connecting the points

Since we are looking for

$y≥0$ , the values are on or above the

$x$ axis

Hence,

$x≤-2.766$ and

$x≥1.266$

$x≤-2.766$ and

$x≥1.266$

Question 3 of 4

3. Question
Graph the inequality:

$y$ $>$ $x^2-3x-4$
- 1.
- 2.
- 3.
- 4.
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Remember the following notations when graphing inequalities.

Symbol Solid / Dotted

$<$ Dotted Line

$>$ Dotted Line

$≤$ Solid Line

$≥$ Solid Line

First, equate the function to $0$ and solve for $x$ by factoring

$y$ $>$ $x^2-3x-4$

$0$ $=$ $x^2-3x-4$

$x^2-3x-4$ $=$ $0$

$(x-4)(x+1)$ $=$ $0$

$x-4$ $=$ $0$

$x-4$ $+4$ $=$ $0$ $+4$

$x$ $=$ $4$

$x+1$ $=$ $0$

$x+1$ $-1$ $=$ $0$ $-1$

$x$ $=$ $-1$

Mark these $2$ points on the $x$ axis

Next, find the $y$ intercept by substituting $x=0$

$y$ $=$ $x^2-3x-4$

$y$ $=$ $(0)^2-3(0)-4$ Substitute $x=0$

$y$ $=$ $0-0-4$

$y$ $=$ $-4$

Mark this on the $y$ axis

Now, connect the points to form a parabola

Remember to use a dotted line because of the $>$ sign

To determine which region to shade, test the origin by substituting $(0,0)$ to the original function

$x=0$

$y=0$

$y$ $>$ $x^2-3x-4$

$0$ $>$ $(0)^2-3(0)-4$ Substitute values

$0$ $>$ $0-0-4$

$0$ $>$ $-4$

This is true, which means the region that includes the origin must be shaded

Question 4 of 4

Graph the system of inequalities:

$y≥x^2-4x+3$

$y$

$<$

$4-x^2$

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Remember the following notations when graphing inequalities.

Symbol	Solid / Dotted
$<$	Dotted Line
$>$	Dotted Line
$≤$	Solid Line
$≥$	Solid Line

Axis of Symmetry

$x=\frac{-\color{#9a00c7}{b}}{2\color{#00880A}{a}}$

First, graph the first inequality

Start by equating the function to

$0$ and solving for

$x$ by factoring

$x^2-4x+3$

$=$

$0$

$(x-3)(x+1)$

$=$

$0$

$x-3$	$=$	$0$
$x-3$ $+3$	$=$	$0$ $+3$
$x$	$=$	$3$

$x+1$	$=$	$0$
$x+1$ $-1$	$=$	$0$ $-1$
$x$	$=$	$-1$

Mark these

$2$ points on the

$x$ axis

Next, find the axis of symmetry

$y=x^2-4x+3$

$a=1$ $b=-4$ $c=3$

$x$	$=$	$\frac{-\color{#9a00c7}{b}}{2\color{#00880A}{a}}$	Axis of Symmetry

$x$	$=$	$\frac{-\color{#9a00c7}{(-4)}}{2\color{#00880A}{(1)}}$	Substitute values

$x$	$=$	$4/2$

$x$	$=$	$2$

Substitute

$x=2$ to the equation to find the value of

$y$ for the vertex

$y$	$=$	$x^2-4x+3$
$y$	$=$	$2^2-4(2)+3$	Substitute $x=2$
$y$	$=$	$4-8+3$
$y$	$=$	$-1$

This means that the vertex is at

$(2,-1)$

Find the

$y$ intercept by substituting

$x=0$

$y$	$=$	$x^2-4x+3$
$y$	$=$	$0^2-4(0)+3$	Substitute $x=0$
$y$	$=$	$0-0+3$
$y$	$=$	$3$

Now, connect the points to form a parabola

Remember to use a solid line because of the

$≥$ sign

To determine which region to shade, test a point by substituting

$(2,0)$ to the original function

$x=2$

$y=0$

$y$	$≥$	$x^2-4x+3$
$0$	$≥$	$2^2-4(2)+3$	Substitute values
$0$	$≥$	$4-8+3$
$0$	$≥$	$-1$

This is true, which means the region that covers

$(2,0)$ must be shaded

This time, graph the second inequality

Start by equating the function to

$0$ and solving for

$x$ by factoring

$4-x^2$	$=$	$0$
$(2-x)(2+x)$	$=$	$0$

$2-x$	$=$	$0$
$2-x$ $+x$	$=$	$0$ $+x$
$2$	$=$	$x$
$x$	$=$	$2$

$2+x$	$=$	$0$
$2+x$ $-x$	$=$	$0$ $-x$
$2$	$=$	$-x$
$x$	$=$	$-2$

Mark these

$2$ points on the

$x$ axis

Next, find the axis of symmetry

$y=4-x^2$

$y=-x^2+4$

$a=-1$ $b=0$ $c=4$

$x$	$=$	$\frac{-\color{#9a00c7}{b}}{2\color{#00880A}{a}}$	Axis of Symmetry

$x$	$=$	$\frac{-\color{#9a00c7}{0}}{2\color{#00880A}{(-1)}}$	Substitute values

$x$	$=$	$0$

The axis of symmetry is at

$x=0$ or the

$y$ axis

Substitute

$x=0$ to the equation to find the value of

$y$ for the vertex

$y$	$=$	$4-x^2$
$y$	$=$	$4-0^2$	Substitute $x=0$
$y$	$=$	$4$

This means that the vertex is at

$(0,4)$

Since this point lies on the

$y$ axis, it is also the

$y$ intercept

Now, connect the points to form a parabola

Remember to use a dotted line because of the

$<$ sign

To determine which region to shade, test the origin by substituting

$(0,0)$ to the original function

$x=0$

$y=0$

$y$	$<$	$4-x^2$
$0$	$<$	$4-0^2$	Substitute values
$0$	$<$	$4$

This is true, which means the region that covers

$(0,0)$ must be shaded

Finally, highlight the overlapping region of the two inequalities

Quadratic Inequalities 2

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Quiz summary

Information

1. Question

Hint

Excellent!

2. Question

Hint

Nice Job!

The Quadratic Formula

3. Question

Hint

Well Done!

4. Question

Hint

Exceptional!

Axis of Symmetry

Quizzes

$3 x^{2} - 3$ $3x^2-3$	$=$ $=$	$0$ $0$
$3 (x^{2} - 1)$ $3(x^2-1)$	$=$ $=$	$0$ $0$	Factor out $3$ $3$
$3 (x + 1) (x - 1)$ $3(x+1)(x-1)$	$=$ $=$	$0$ $0$	Difference of two squares

$x + 1$ $x+1$	$=$ $=$	$0$ $0$
$x + 1$ $x+1$ $- 1$ $-1$	$=$ $=$	$0$ $0$ $- 1$ $-1$
$x$ $x$	$=$ $=$	$- 1$ $-1$

$x - 1$ $x-1$	$=$ $=$	$0$ $0$
$x - 1$ $x-1$ $+ 1$ $+1$	$=$ $=$	$0$ $0$ $+ 1$ $+1$
$x$ $x$	$=$ $=$	$1$ $1$

$y$ $y$	$=$ $=$	$3 x^{2} - 3$ $3x^2-3$
$y$ $y$	$=$ $=$	$3 {(0)}^{2} - 3$ $3(0)^2-3$	Substitute $x=0$
$y$	$=$	$0-3$
$y$	$=$	$-3$

$y$	$=$	$x^2-3x-4$
$y$	$=$	$(0)^2-3(0)-4$	Substitute $x=0$
$y$	$=$	$0-0-4$
$y$	$=$	$-4$