Abeka Grade 9 Algebra 1 Test 9 Flashcards

Multiple Choice Overview

• For the equation (9x²-3x+1=0), the discriminant (b²-4ac) results in (-27), indicating 0 real solutions.
• The zeros of the quadratic (y=x²+6x-27) are found to be (3,0) and (-9,0) by factoring as ((x-3)(x+9)=0).

Vertex Identification

• The vertex for (y=(x-1)²-5) is determined to be (1,-5) using the vertex form (y=(x-h)²+k).
• For the equation (y=x²+4x+3), the vertex is calculated as (-2,-1) through the formula (x=-b/2a).

Distance Calculation

• The distance between points (1,3) and (4,-1) is found using the distance formula, resulting in 5.

Geometry Problem

• Given an 18-inch wide screen with a diagonal of 20 inches, the height is approximately 9 inches after applying the Pythagorean theorem.

Polynomial Factoring

• The expression (8a²b²+10ab³c) factors to (2ab²(4a+5bc)).
• The polynomial (2x³+2x²+x+1) can be factored to ((x+1)(2x²+1)).

Simplification

• The expression (\sqrt{12x⁵}) simplifies to (2x²\sqrt{3x}).
• The cube root of (-8) evaluates to 2.

Completing Operations

• The expression ((\sqrt{a}+2)(\sqrt{a}+3)) expands to (a+5\sqrt{a}+6).
• The simplification of (\frac{4y}{\sqrt{2y}}) leads to (2\sqrt{2y}).

Rationalizing Denominators

• The expression (\frac{5}{\sqrt{7}-1}) simplifies to (\frac{5\sqrt{7}+5}{6}) through rationalization.

Solving Equations

• For the equation (5x²-3=17), the solutions are (x=±2) after rearranging and taking the square root.
• The quadratic (x²+5x-3=0) yields roots of (x=\frac{1}{2}) or (x=-3) using the quadratic formula.

Additional Equations

• The equation (x²-x=6) is factored to find solutions (x=-2) or (x=3).
• For (x²-2x-2=0), the solutions are (x=1±\sqrt{3}), derived from completing the square.
• The equation (\sqrt{x²-4x+13}=x+7) results in (x=-2), confirmed by verification of the solution.