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Solve for x. Enter the solutions from least to greatest. {:[x^(2)-12 x+27=0],[" lesser "x=◻],[" greater "x=◻]:}

Solve for x x . Enter the solutions from least to greatest.\newlinex2−12x+27=0 lesser x=□ greater x=□ \begin{array}{l} x^{2}-12 x+27=0 \\ \text { lesser } x=\square \\ \text { greater } x=\square \end{array}

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Q. Solve for x x . Enter the solutions from least to greatest.\newlinex2−12x+27=0 lesser x=□ greater x=□ \begin{array}{l} x^{2}-12 x+27=0 \\ \text { lesser } x=\square \\ \text { greater } x=\square \end{array}
  1. Identify quadratic equation: Identify the quadratic equation to be solved.\newlineThe given quadratic equation is x2−12x+27=0x^2 - 12x + 27 = 0. We need to find two numbers that multiply to 2727 and add up to −12-12.
  2. Factor the equation: Factor the quadratic equation.\newlineThe two numbers that multiply to 2727 and add up to −12-12 are −9-9 and −3-3 because (−9)×(−3)=27(-9) \times (-3) = 27 and (−9)+(−3)=−12(-9) + (-3) = -12. So, we can write the equation as (x−9)(x−3)=0(x - 9)(x - 3) = 0.
  3. Solve for x (first factor): Solve for x by setting each factor equal to zero.\newlineFirst, set the first factor equal to zero: x−9=0x - 9 = 0. Then, solve for x by adding 99 to both sides: x=9x = 9.
  4. Solve for x (second factor): Solve for x using the second factor.\newlineNow, set the second factor equal to zero: x−3=0x - 3 = 0. Then, solve for x by adding 33 to both sides: x=3x = 3.
  5. Write the solutions: Write the solutions in ascending order.\newlineThe solutions are x=3x = 3 and x=9x = 9. Since 33 is less than 99, we write them as lesser x=3x = 3 and greater x=9x = 9.

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