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Grade 9 · unit review

Review: Algebra1: Quadratic Solution Structures

Quadratic equations can be solved by revealing factors, making a square, or undoing a square. Each method exposes structure, and all valid roots must be retained and checked in the original equation.

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Use these 4 new cases to check your reasoning. Choices are temporary and are not saved to an account, sent to staff or recorded as school grades. Download your current review choices before leaving or reloading.

Unit review questions
1. Factor x²−15x+56.
Need a hint?
  • Factoring a monic quadratic into two linear factors requires numbers matching both the linear coefficient’s sum and the constant term’s product.
  • Matching the product alone is insufficient.
2. Solve (x−7)(x+8)=0.
Need a hint?
  • A product of real factors equals zero only if at least one factor is zero.
  • The zero-product property applies when the whole product equals zero, not an arbitrary nonzero value.
3. What constant makes x²+14x+___ a perfect-square trinomial?
Need a hint?
  • A square (x+a)² expands to x²+2ax+a².
  • When completing an equation’s square, add the constant to both sides.
4. Solve (x−7)²=64.
Need a hint?
  • A squared expression equal to a positive square can have either positive or negative square-root value.
  • The principal square-root symbol itself is nonnegative, but solving z²=b requires z=±√b for positive b.

The download contains your current choices and any checked feedback. Keep the separate transfer-task work in your own document; downloading its instructions does not mark it complete.

Use the skills in a new task

Create a four-part mathematical solution sheet for these supplied cases. 1. Factor x²−19x+90. Candidate answers: (A) (x+9)(x+10) (B) (x−19)² (C) (x−9)(x−10) (D) (x−90)(x−1). 2. Solve (x−9)(x+10)=0. Candidate answers: (A) x=-1 (B) x=9 only (C) x=9 or x=−10 (D) x=−9 or x=10. 3. What constant makes x²+18x+___ a perfect-square trinomial? Candidate answers: (A) 162 (B) 80 (C) 81 (D) 82. 4. Solve (x−9)²=100. Candidate answers: (A) x=-1 or x=19 (B) x=-91 or x=109 (C) x=109 (D) x=19 only.

Work on paper or in your own document. Explain the choices you made.

  • State each requested quantity or relationship and its input restrictions.
  • Show intermediate calculations or the applicable theorem with matching correspondences.
  • Give all four results, including exact values or requested rounding and units.
  • Check a result by substitution, reversal, dimensions, a representation or a counterexample as appropriate.
Compare with a model response

1. (x−9)(x−10). The constant term requires two numbers whose product is 90. Their sum must be −19; −9 and −10 meet both conditions. 2. x=9 or x=−10. A product is zero only when at least one factor is zero. Solve x−9=0 and x+10=0 separately; both solutions satisfy the product equation. 3. 81. Half the x coefficient is 9; square it to get 81. The trinomial becomes (x+9)². Adding it to an equation requires adding the same amount on both sides. 4. x=-1 or x=19. Take both square-root possibilities: x−9=±10. Add 9 to each: x=-1 or 19. A square equation can have two real solutions.

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