Solve the Quadratic Inequality: x²-6x+9>0

Quadratic Inequalities with Perfect Square Trinomials

Solve the following equation:

x2−6x+9>0 x^2-6x+9>0

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1

Understand the problem

Solve the following equation:

x2−6x+9>0 x^2-6x+9>0

2

Step-by-step solution

To solve the quadratic inequality x2−6x+9>0 x^2 - 6x + 9 > 0 , we first rewrite the quadratic expression in a recognizable form:

The expression can be rewritten as:

x2−6x+9=(x−3)2 x^2 - 6x + 9 = (x-3)^2

This is a perfect square trinomial, where (x−3)2>0 (x-3)^2 > 0 . We know that a square of a real number is only greater than zero when the number itself is not zero.

Thus, (x−3)2>0 (x-3)^2 > 0 implies x−3≠0 x - 3 \neq 0 , meaning x≠3 x \neq 3 .

Consequently, the inequality x2−6x+9>0 x^2 - 6x + 9 > 0 holds true for all x x except x=3 x = 3 .

Therefore, the solution to the inequality is:

3≠x 3 \neq x

3

Final Answer

3≠x 3 ≠ x

Key Points to Remember

Essential concepts to master this topic
  • Perfect Square Recognition: x2−6x+9=(x−3)2 x^2 - 6x + 9 = (x-3)^2 by factoring pattern
  • Inequality Rule: (x−3)2>0 (x-3)^2 > 0 when x ≠ 3 since squares are non-negative
  • Solution Check: Test x = 2: (2−3)2=1>0 (2-3)^2 = 1 > 0 ✓

Common Mistakes

Avoid these frequent errors
  • Setting the quadratic equal to zero instead of analyzing the inequality
    Don't solve x2−6x+9=0 x^2 - 6x + 9 = 0 and stop there = only finding where expression equals zero! This misses that we need values where it's greater than zero. Always recognize that (x−3)2>0 (x-3)^2 > 0 for all x except x = 3.

Practice Quiz

Test your knowledge with interactive questions

Solve the following equation:

\( x^2+4>0 \)

FAQ

Everything you need to know about this question

Why isn't the answer 'for all values of x'?

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Because when x = 3, we get (3−3)2=0 (3-3)^2 = 0 , and we need the expression to be greater than 0, not equal to 0. So x = 3 must be excluded from the solution.

How do I recognize a perfect square trinomial?

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Look for the pattern a2−2ab+b2=(a−b)2 a^2 - 2ab + b^2 = (a-b)^2 . Here, x2−6x+9 x^2 - 6x + 9 has first term x², last term 3² = 9, and middle term -2(x)(3) = -6x.

What does x≠3 x \neq 3 mean in interval notation?

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It means (−∞,3)∪(3,∞) (-\infty, 3) \cup (3, \infty) . This represents all real numbers except 3. Notice the parentheses around 3 show it's excluded.

Could this inequality ever have no solution?

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Not for perfect squares! Since (x−3)2≥0 (x-3)^2 \geq 0 always, it's greater than 0 everywhere except where it equals 0. Perfect squares are never negative.

How do I check my answer?

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Pick any value except x = 3 and substitute it. Try x = 0: 02−6(0)+9=9>0 0^2 - 6(0) + 9 = 9 > 0 ✓. Try x = 5: 25−30+9=4>0 25 - 30 + 9 = 4 > 0 ✓.

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