Quadratic Inequalities 1

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Year 11

Quadratic Polynomial

Quadratic Inequalities

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. $- 3 \leq x \leq 4$ $-3≤x≤4$
- 4. $x \leq 0$ $x≤0$ and $x \geq 4$ $x≥4$
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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

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

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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

$x$ 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. $x < 4$ $x<4$ and $x < - \frac{1}{2}$ $x<-1/2$
- 2. $4 > x > - \frac{1}{2}$ $4>x> -1/2$
- 3. `-4
- 4. $x$ $x$ $<$ $<$ $- \frac{1}{2}$ $-1/2$ and $x$ $x$ $>$ $>$ $4$ $4$
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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 $x$ $x$ 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 4

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. $\frac{2}{5} \leq x \leq - 1$ $2/5≤x≤-1$
- 4. $1 \leq x \leq \frac{2}{5}$ $1≤x≤2/5$
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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, $\frac{2}{5} \leq x \leq 1$ $2/5≤x≤1$

$\frac{2}{5} \leq x \leq 1$ $2/5≤x≤1$

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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!

Quizzes

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

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

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

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

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

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

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

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

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