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Solve the Inequality
(Quadratic Inequalities)
Question 1 of 3
Enter the two boundary values (the roots), then select the correct inequality direction.
Solution:
Step 1: The expression is already in standard form
\[x^2 - 4x + 3 > 0\]
Step 2: Set equal to zero and solve for the roots
Replace \(>\) with \(=\) to find where the expression equals zero:
\[x^2 - 4x + 3 = 0\]
Factorise:
\[(x-1)(x-3) = 0\]
\(x = 1\) or \(x = 3\)
Step 3: Test a point in each region to find where the expression is positive
The roots are \(x=1\) and \(x=3\), which create three regions. Test one point from each:
Left region — test \(x = 0\):
\[(0)^2 - 4(0) + 3 = 3 \Rightarrow \text{positive}\]
Middle region — test \(x = 2\):
\[(2)^2 - 4(2) + 3 = 4 - 8 + 3 = -1 \Rightarrow \text{negative}\]
Right region — test \(x = 4\):
\[(4)^2 - 4(4) + 3 = 16 - 16 + 3 = 3 \Rightarrow \text{positive}\]
We need the expression to be positive (> 0), so the answer is the left and right regions.
Step 4: Write the answer
\[x 3\]
Number line:
Enter the two boundary values (the roots), then select the correct inequality direction.
Solution:
Step 1: The expression is already in standard form
\[x^2 - 5x + 6 Step 2: Set equal to zero and solve for the roots
Replace \(<\) with \(=\) to find where the expression equals zero:
\[x^2 - 5x + 6 = 0\]
Factorise:
\[(x-2)(x-3) = 0\]
\(x = 2\) or \(x = 3\)
Step 3: Test a point in each region to find where the expression is negative
The roots are \(x=2\) and \(x=3\), which create three regions. Test one point from each:
Left region — test \(x = 0\):
\[(0)^2 - 5(0) + 6 = 6 \Rightarrow \text{positive}\]
Middle region — test \(x = 2.5\):
\[(2.5)^2 - 5(2.5) + 6 = 6.25 - 12.5 + 6 = -0.25 \Rightarrow \text{negative}\]
Right region — test \(x = 4\):
\[(4)^2 - 5(4) + 6 = 16 - 20 + 6 = 2 \Rightarrow \text{positive}\]
We need the expression to be negative (< 0), so the answer is the middle region.
Step 4: Write the answer
\[2 Which can also be written as: \(x > 2\) and \(x
Number line:
Enter the two boundary values (the roots), then select the correct inequality direction.
Solution:
Step 1: The expression is already in standard form
\[x^2 - 7x + 12 > 0\]
Step 2: Set equal to zero and solve for the roots
Replace \(>\) with \(=\) to find where the expression equals zero:
\[x^2 - 7x + 12 = 0\]
Factorise:
\[(x-3)(x-4) = 0\]
\(x = 3\) or \(x = 4\)
Step 3: Test a point in each region to find where the expression is positive
The roots are \(x=3\) and \(x=4\), which create three regions. Test one point from each:
Left region — test \(x = 0\):
\[(0)^2 - 7(0) + 12 = 12 \Rightarrow \text{positive}\]
Middle region — test \(x = 3.5\):
\[(3.5)^2 - 7(3.5) + 12 = 12.25 - 24.5 + 12 = -0.25 \Rightarrow \text{negative}\]
Right region — test \(x = 5\):
\[(5)^2 - 7(5) + 12 = 25 - 35 + 12 = 2 \Rightarrow \text{positive}\]
We need the expression to be positive (> 0), so the answer is the left and right regions.
Step 4: Write the answer
\[x 4\]
Number line:
