Quadratic Inequalities

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Question 1 of 8

1. Question
Solve for $x$ $x$

$x^{2} - x - 12 \leq 0$ $x^2-x-12≤0$
- 1. $x \leq 0$ $x≤0$ and $x \geq 4$ $x≥4$
- 2. $- 3 \leq x \leq 4$ $-3≤x≤4$
- 3. $x \leq - 3$ $x≤-3$ and $x \geq 4$ $x≥4$
- 4. $3 \leq x \leq 4$ $3≤x≤4$
Hint

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Incorrect

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Representing Inequalities on the Number Line

Greater than ( $>$ $>$ )

Greater than or equal ( $\geq$ $≥$ )

Less than ( $<$ $<$ )

Less than or equal ( $\leq$ $≤$ )

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

$x^{2} - x - 12$ $x^2-x-12$ $\leq$ $≤$ $0$ $0$

$x^{2} - x - 12$ $x^2-x-12$ $=$ $=$ $0$ $0$

$(x + 3) (x - 4)$ $(x+3)(x-4)$ $=$ $=$ $0$ $0$

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

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

$x$ $x$ $=$ $=$ $- 2$ $-2$

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

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

$x$ $x$ $=$ $=$ $4$ $4$

Mark these $2$ $2$ points on a number plane. Use filled dots since the sign used is $\leq$ $≤$

Next, test a point to determine which part of the number line is covered by $x$ $x$

Try $x = - 5$ $x=-5$

$x^{2} - x - 12$ $x^2-x-12$ $\leq$ $≤$ $0$ $0$

${(- 5)}^{2} - (- 5) - 12$ $(-5)^2-(-5)-12$ $\leq$ $≤$ $0$ $0$ Substitute $x = - 5$ $x=-5$

$25 + 5 - 12$ $25+5-12$ $\leq$ $≤$ $0$ $0$

$18$ $18$ $\leq$ $≤$ $0$ $0$

This is not true, which means $- 3 \leq x \leq 4$ $-3≤x≤4$

$- 3 \leq x \leq 4$ $-3≤x≤4$

Question 2 of 8

Solve for

x

$x$

4 + 3 x - x^{2} \geq 0

$4+3x-x^2≥0$

1. $- 4 \leq x \leq 1$ $-4≤x≤1$
2. $1 \leq x \leq 4$ $1≤x≤4$
3. $x \leq - 1$ $x≤-1$ and $x \geq 4$ $x≥4$
4. $- 1 \leq x \leq 4$ $-1≤x≤4$

Question 3 of 8

3. Question
Solve for $x$ $x$

$2 x^{2} > 7 x + 4$ $2x^2>7x+4$
- 1. $- 4 < x < \frac{1}{2}$ $-4<x<1/2$
- 2. $x$ $x$ $<$ $<$ $- \frac{1}{2}$ $-1/2$ and $x$ $x$ $>$ $>$ $4$ $4$
- 3. $4 > x > - \frac{1}{2}$ $4>x> -1/2$
- 4. $x < 4$ $x<4$ and $x < - \frac{1}{2}$ $x<-1/2$
Hint

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Correct

Keep Going!

Incorrect

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The cross method is a factorisation method used for quadratics.

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

$2 x^{2}$ $2x^2$ $=$ $=$ $7 x + 4$ $7x+4$

$2 x^{2} - 7 x - 4$ $2x^2-7x-4$ $=$ $=$ $0$ $0$ Convert to standard form

$2 x^{2} - 7 x - 4$ $2x^2-7x-4$ $=$ $=$ $0$ $0$

$(2 x + 1) (x - 4)$ $(2x+1)(x-4)$ $=$ $=$ $0$ $0$

$2 x + 1$ $2x+1$ $=$ $=$ $0$ $0$

$2 x + 1$ $2x+1$ $- 1$ $-1$ $=$ $=$ $0$ $0$ $- 1$ $-1$

$2 x$ $2x$ $\div 2$ $divide2$ $=$ $=$ $- 1$ $-1$ $\div 2$ $divide2$

$x$ $x$ $=$ $=$ $- \frac{1}{2}$ $-1/2$

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

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

$x$ $x$ $=$ $=$ $4$ $4$

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

$y$ $y$ $=$ $=$ $2 x^{2} - 7 x - 4$ $2x^2-7x-4$

$y$ $y$ $=$ $=$ $2 {(0)}^{2} - 7 (0) - 4$ $2(0)^2-7(0)-4$ Substitute $x = 0$ $x=0$

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

Mark this point on the $y$ $y$ axis.

Form a parabola by connecting the points

Since we are looking for $y$ $y$ $>$ $>$ $0$ $0$ , the values are above the $x$ $x$ axis

Hence, $x$ $x$ $<$ $<$ $- \frac{1}{2}$ $-1/2$ and $x$ $x$ $>$ $>$ $4$ $4$

$x$ $x$ $<$ $<$ $- \frac{1}{2}$ $-1/2$ and $x$ $x$ $>$ $>$ $4$ $4$
Question 4 of 8

4. Question
Solve for $x$ $x$

$- 5 x^{2} + 7 x - 2 \geq 0$ $-5x^2+7x-2≥0$
- 1. $- \frac{2}{5} \leq x \leq 1$ $-2/5≤x≤1$
- 2. $\frac{2}{5} \leq x \leq - 1$ $2/5≤x≤-1$
- 3. $1 \leq x \leq \frac{2}{5}$ $1≤x≤2/5$
- 4. $\frac{2}{5} \leq x \leq 1$ $2/5≤x≤1$
Hint

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Correct

Fantastic!

Incorrect

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The cross method is a factorisation method used for quadratics.

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

$- 5 x^{2} + 7 x - 2$ $-5x^2+7x-2$ $\geq$ $≥$ $0$ $0$

$- 5 x^{2} + 7 x - 2$ $-5x^2+7x-2$ $=$ $=$ $0$ $0$

$(5 x - 2) (x - 1)$ $(5x-2)(x-1)$ $=$ $=$ $0$ $0$

$5 x - 2$ $5x-2$ $=$ $=$ $0$ $0$

$5 x - 2$ $5x-2$ $+ 2$ $+2$ $=$ $=$ $0$ $0$ $+ 2$ $+2$

$5 x$ $5x$ $\div 5$ $divide5$ $=$ $=$ $2$ $2$ $\div 5$ $divide5$

$x$ $x$ $=$ $=$ $\frac{2}{5}$ $2/5$

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

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

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

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

$y$ $y$ $=$ $=$ $- 5 x^{2} + 7 x - 2$ $-5x^2+7x-2$

$y$ $y$ $=$ $=$ $- 5 {(0)}^{2} + 7 (0) - 2$ $-5(0)^2+7(0)-2$ Substitute $x = 0$ $x=0$

$y$ $y$ $=$ $=$ $- 2$ $-2$

Mark this point on the $y$ $y$ axis.

Form a parabola by connecting the points

Since we are looking for $y \geq 0$ $y≥0$ , the values are above the $x$ $x$ axis

Hence, $\frac{2}{5} \leq x \leq 1$ $2/5≤x≤1$

$\frac{2}{5} \leq x \leq 1$ $2/5≤x≤1$
Question 5 of 8

5. 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. $- 3$ $-3$ $<$ $<$ $x$ $x$ $<$ $<$ $3$ $3$
- 3. $x$ $x$ $<$ $<$ $- 1$ $-1$ and $x$ $x$ $>$ $>$ $1$ $1$
- 4. $- 1$ $-1$ $<$ $<$ $x$ $x$ $<$ $<$ $1$ $1$
Hint

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Correct

Excellent!

Incorrect

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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 $x = 0$ $x=0$

$y$ $y$ $=$ $=$ $0 - 3$ $0-3$

$y$ $y$ $=$ $=$ $- 3$ $-3$

Mark this point on the $y$ $y$ axis.

Form a parabola by connecting the points

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

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

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

Question 6 of 8

Solve for

x

$x$ :

2 x^{2} + 3 x - 7 \geq 0

$2x^2+3x-7≥0$

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

Incorrect

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The Quadratic Formula

x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

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

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

x

$x$ using the Quadratic Formula

2 x^{2} + 3 x - 7 = 0

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

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

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

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

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

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

$x$ axis

Next, find the

y

$y$ intercept by substituting

x = 0

$x=0$

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

Mark this on the

y

$y$ axis

Form a parabola by connecting the points

Since we are looking for

y \geq 0

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

x

$x$ axis

Hence,

x \leq - 2.766

$x≤-2.766$ and

x \geq 1.266

$x≥1.266$

x \leq - 2.766

$x≤-2.766$ and

x \geq 1.266

$x≥1.266$

Question 7 of 8

7. Question
Graph the inequality:

$y$ $y$ $>$ $>$ $x^{2} - 3 x - 4$ $x^2-3x-4$
- 1.
- 2.
- 3.
- 4.
Hint

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Correct

Well Done!

Incorrect

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Remember the following notations when graphing inequalities.

Symbol Solid / Dotted

$<$ $<$ Dotted Line

$>$ $>$ Dotted Line

$\leq$ $≤$ Solid Line

$\geq$ $≥$ Solid Line

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

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

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

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

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

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

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

$x$ $x$ $=$ $=$ $4$ $4$

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

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

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

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

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

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

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

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

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

Mark this on the $y$ $y$ 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 $(0, 0)$ $(0,0)$ to the original function

$x = 0$ $x=0$

$y = 0$ $y=0$

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

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

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

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

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

Question 8 of 8

Graph the system of inequalities:

y \geq x^{2} - 4 x + 3

$y≥x^2-4x+3$

y

$y$

<

$<$

4 - x^{2}

$4-x^2$

Incorrect

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Remember the following notations when graphing inequalities.

Symbol	Solid / Dotted
$<$ $<$	Dotted Line
$>$ $>$	Dotted Line
$\leq$ $≤$	Solid Line
$\geq$ $≥$	Solid Line

Axis of Symmetry

x = \frac{- b}{2 a}

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

First, graph the first inequality

Start by equating the function to

0

$0$ and solving for

x

$x$ by factoring

x^{2} - 4 x + 3

$x^2-4x+3$

=

$=$

0

$0$

(x - 3) (x + 1)

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

=

$=$

0

$0$

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

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

Mark these

2

$2$ points on the

x

$x$ axis

Next, find the axis of symmetry

y = x^{2} - 4 x + 3

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

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

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

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

$x$ $x$	$=$ $=$	$\frac{4}{2}$ $4/2$

$x$ $x$	$=$ $=$	$2$ $2$

Substitute

x = 2

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

y

$y$ for the vertex

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

This means that the vertex is at

(2, - 1)

$(2,-1)$

Find the

y

$y$ intercept by substituting

x = 0

$x=0$

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

Now, connect the points to form a parabola

Remember to use a solid line because of the

\geq

$≥$ sign

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

(2, 0)

$(2,0)$ to the original function

x = 2

$x=2$

y = 0

$y=0$

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

This is true, which means the region that covers

(2, 0)

$(2,0)$ must be shaded

This time, graph the second inequality

Start by equating the function to

0

$0$ and solving for

x

$x$ by factoring

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

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

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

Mark these

2

$2$ points on the

x

$x$ axis

Next, find the axis of symmetry

y = 4 - x^{2}

$y=4-x^2$

y = - x^{2} + 4

$y=-x^2+4$

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

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

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

$x$ $x$	$=$ $=$	$0$ $0$

The axis of symmetry is at

x = 0

$x=0$ or the

y

$y$ axis

Substitute

x = 0

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

y

$y$ for the vertex

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

This means that the vertex is at

(0, 4)

$(0,4)$

Since this point lies on the

y

$y$ axis, it is also the

y

$y$ 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)

$(0,0)$ to the original function

x = 0

$x=0$

y = 0

$y=0$

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

This is true, which means the region that covers

(0, 0)

$(0,0)$ must be shaded

Finally, highlight the overlapping region of the two inequalities

Quadratic Inequalities

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Quiz summary

Information

1. Question

Hint

Great Work!

Representing Inequalities on the Number Line

2. Question

Hint

Correct!

3. Question

Hint

Keep Going!

4. Question

Hint

Fantastic!

5. Question

Hint

Excellent!

6. Question

Hint

Nice Job!

The Quadratic Formula

7. Question

Hint

Well Done!

8. Question

Hint

Exceptional!

Axis of Symmetry

Quizzes

$x^{2} - x - 12$ $x^2-x-12$	$\leq$ $≤$	$0$ $0$
$x^{2} - x - 12$ $x^2-x-12$	$=$ $=$	$0$ $0$

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

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

$4 + 3 x - x^{2}$ $4+3x-x^2$	$=$ $=$	$0$ $0$
$- x^{2} + 3 x + 4$ $-x^2+3x+4$	$=$ $=$	$0$ $0$	Convert to standard form
$x^{2} - 3 x - 4$ $x^2-3x-4$	$=$ $=$	$0$ $0$	Multiply the function by $- 1$ $-1$
$x^{2} - 3 x - 4$ $x^2-3x-4$	$=$ $=$	$0$ $0$

$2 x^{2}$ $2x^2$	$=$ $=$	$7 x + 4$ $7x+4$
$2 x^{2} - 7 x - 4$ $2x^2-7x-4$	$=$ $=$	$0$ $0$	Convert to standard form
$2 x^{2} - 7 x - 4$ $2x^2-7x-4$	$=$ $=$	$0$ $0$

$2 x + 1$ $2x+1$	$=$ $=$	$0$ $0$
$2 x + 1$ $2x+1$ $- 1$ $-1$	$=$ $=$	$0$ $0$ $- 1$ $-1$
$2 x$ $2x$ $\div 2$ $divide2$	$=$ $=$	$- 1$ $-1$ $\div 2$ $divide2$

$x$ $x$	$=$ $=$	$- \frac{1}{2}$ $-1/2$

$- 5 x^{2} + 7 x - 2$ $-5x^2+7x-2$	$\geq$ $≥$	$0$ $0$
$- 5 x^{2} + 7 x - 2$ $-5x^2+7x-2$	$=$ $=$	$0$ $0$

$5 x - 2$ $5x-2$	$=$ $=$	$0$ $0$
$5 x - 2$ $5x-2$ $+ 2$ $+2$	$=$ $=$	$0$ $0$ $+ 2$ $+2$
$5 x$ $5x$ $\div 5$ $divide5$	$=$ $=$	$2$ $2$ $\div 5$ $divide5$

$x$ $x$	$=$ $=$	$\frac{2}{5}$ $2/5$

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

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