Let’s solve the inequality step by step:
We are given:
> (x – 3)² + 2 < |x – 7|
First, expand the square:
> (x – 3)² = x² – 6x + 9
So the left side becomes:
> x² – 6x + 9 + 2 = x² – 6x + 11
Now the inequality is:
> x² – 6x + 11 < |x – 7|
This is a quadratic on the left and an absolute value on the right. To solve this, we need to consider two cases based on the definition of absolute value.
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Case 1: x – 7 ≥ 0 → x ≥ 7
Then |x – 7| = x – 7
Inequality becomes:
> x² – 6x + 11 < x – 7
Bring all terms to one side:
> x² – 6x + 11 – x + 7 < 0
> x² – 7x + 18 < 0
Now check if this quadratic can be negative.
Discriminant D = (-7)² – 4(1)(18) = 49 – 72 = -23 → negative
Since discriminant is negative and leading coefficient is positive, the quadratic x² – 7x + 18 is always
positive for all real x.
So, x² – 7x + 18 < 0 has
no solution.
→ No solutions in Case 1 (x ≥ 7)
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Case 2: x – 7 < 0 → x < 7
Then |x – 7| = -(x – 7) = -x + 7
Inequality becomes:
> x² – 6x + 11 < -x + 7
Bring all terms to one side:
> x² – 6x + 11 + x – 7 < 0
> x² – 5x + 4 < 0
Factor the quadratic:
> (x – 1)(x – 4) < 0
This inequality holds when x is between the roots:
> 1 < x < 4
But remember — this case only applies when x < 7. Since 1 < x < 4 is already within x < 7, we keep it.
→ Solutions from Case 2: 1 < x < 4
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Now combine both cases:
- Case 1: no solution
- Case 2: 1 < x < 4
Final Answer:
1 < x < 4
Parent Tip: Review the logic above to help your child master the concept of absolute value questions.
