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Quadratic Inequalities 1

Quadratic Inequalities 1

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Question 1 of 4

1. Question
Solve for $x$ $x$

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

2. Question

Solve for

x

$x$

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

$4+3x-x^2≥0$

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

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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

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

(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, substitute

x = 0

$x=0$ to the function to get the

y

$y$ intercept

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

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

x

$x$ axis

This means that

x

$x$ is greater than or equal to

- 1

$-1$ and less than or equal to

4

$4$

- 1 \leq x \leq 4

$-1≤x≤4$

Question 3 of 4

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. `-4
- 3. $x<4$ and $x<-1/2$
- 4. $x$ $<$ $-1/2$ and $x$ $>$ $4$
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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 $x$ values using cross method

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

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

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

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

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

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

$2x$ $divide2$ $=$ $-1$ $divide2$

$x$ $=$ $-1/2$

$x-4$ $=$ $0$

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

$x$ $=$ $4$

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

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

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

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

$y$ $=$ $-4$

Mark this point on the $y$ axis.

Form a parabola by connecting the points

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

Hence, $x$ $<$ $-1/2$ and $x$ $>$ $4$

$x$ $<$ $-1/2$ and $x$ $>$ $4$
Question 4 of 4

4. Question
Solve for $x$

$-5x^2+7x-2≥0$
- 1. $-2/5≤x≤1$
- 2. $1≤x≤2/5$
- 3. $2/5≤x≤-1$
- 4. $2/5≤x≤1$
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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 $x$ values using cross method

$-5x^2+7x-2$ $≥$ $0$

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

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

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

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

$5x$ $divide5$ $=$ $2$ $divide5$

$x$ $=$ $2/5$

$x-1$ $=$ $0$

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

$x$ $=$ $1$

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

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

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

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

$y$ $=$ $-2$

Mark this point on the $y$ axis.

Form a parabola by connecting the points

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

Hence, $2/5≤x≤1$

$2/5≤x≤1$

Quizzes

Sum & Product of Roots 1
Sum & Product of Roots 2
Sum & Product of Roots 3
Sum & Product of Roots 4
Solving Equations by Factoring 1
Solving Equations Using the Quadratic Formula
Completing the Square 1
Completing the Square 2
Intro to Quadratic Functions (Parabolas) 1
Intro to Quadratic Functions (Parabolas) 2
Intro to Quadratic Functions (Parabolas) 3
Graph Quadratic Functions in Standard Form 1
Graph Quadratic Functions in Standard Form 2
Graph Quadratic Functions by Completing the Square
Graph Quadratic Functions in Vertex Form
Write a Quadratic Equation from the Graph
Write a Quadratic Equation Given the Vertex and Another Point
Quadratic Inequalities 1
Quadratic Inequalities 2
Quadratics Word Problems 1
Quadratics Word Problems 2
Quadratic Identities
Graphing Quadratics Using the Discriminant
Positive and Negative Definite
Applications of the Discriminant 1
Applications of the Discriminant 2
Solving Reducible Equations

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