Math Quadratics Midterm Review Key

Algebra 2 Midterm Review Name KEY Hour Quadratics Factor the following completely. (t^{2}+2t-15) [(t+5)(t-3)] (5v^{2}+4v-1) [(5v-1)(v+1)] (3x^{2}-8x+5) [(3x-5)(x-1)] (3x^{2}y^{4}-81x^{2}y) [3x^{2}y(y^{3}-27)] [3x^{2}y(y-3)(y^{2}+3y+9)] (8p^{3}+1) [(2p+1)(4p^{2}+2p+1)] (16x^{2}+40xy+25y^{2}) [(4x+5y)(4x+5y)] [(4x+5y)^{2}] (4x^{3}+8x^{2}y-5xy^{2}) (x(4x^{2}+8xy-5y^{2})) [x(2x-y)(2x+5y)] (r^{4}-16s^{4}) [(r^{2}+4s^{2})(r^{2}-4s^{2})] [(r^{2}+4s^{2})(r+2s)(r-2s)] (2u^{2}+uv-21v^{2}) [(2u+7)(u-3)] (2(a+b)^{2}+5(a+b)-3) Let (u=a+b) [2u^{2}+5u-3] [(2u-1)(u+3)] [(2(a+b)-1)((a+b)+3)] (2(\frac{x+2}{3})^{2}-5(\frac{x+2}{3})+3) Let (u=\frac{x+2}{3}) [2u^{2}-5u+3] [(2u-1)(u-1)] [(2(\frac{x+2}{3})-1)(\frac{x+2}{3}-1)] Solve by Factoring. ((x+2)(x^{2}-4)=0) [(x+2)(x+2)(x-2)=0] [x=-2,2] (double root) ((x^{2}-1)(x^{2}+3x+2)=\emptyset) [(x+1)(x-1)(x+2)(x+1)=0] [x=-1,1,-2] (y^{4}-5y^{2}+4=0) [(y^{2}-4)(y^{2}-1)=0] [(y+2)(y-2)(y+1)(y-1)=0] [y=-2,2,-1,1] ((3a-4)(a-2)=1) [3a^{2}-6a-4a+8=1] [3a^{2}-10a+7=0] [(3a-7)(a-1)=0] [a=\frac{7}{3},1] (27x^{3}+5=4) [27x^{3}+1=0] [(3x+1)(9x^{2}-3x+1)=0] [x=-\frac{1}{3}, \frac{1\pm i\sqrt{3}}{6}] ((2x-5)^{2}=12) [4x^{2}-20x+25=12] [4x^{2}-20x+13=0] [x=\frac{20\pm\sqrt{400-4(4)(13)}}{8}=\frac{20\pm\sqrt{142}}{8}=\frac{5\pm2\sqrt{3}}{2}] 18. (a^2+1)^2-7(a^2+1)=-10 Let u = a^2+1 [u^2-7u+10=0] [(u-5)(u-2)=0] [u=5] [u=2] [a^2+1=5] [a^2+1=2] [a^2=4] [a^2=1] [a=\pm2] [a=\pm1] 19. 5x+13√x-6=0 Let u = √x [5u^2+13u-6=0] [(5u-2)(u+3)=0] [u=\frac{2}{5}] [u=-3] [√x=\frac{2}{5}] [x=\frac{4}{25}] [√x=-3] (no solution) Solve the following inequalities. 20. x^2+9≤6x [x^2-6x+9≤0] [(x-3)(x-3)≤0] [x=3] (only =0 when x=3) 21. 9-3t-2t^2>0 [(3-2t)(3+t)>0] [t=-3] [t=\frac{3}{2}] 22. z^2+7z+10≥0 [(z+5)(z+2)≥0] [z=-5] [z=-2] Simplify the following expressions. 24. √(-27)-√(-12)+2√75 [3i√3-2i√3+10√3] [10√3+i√3] 25. 4i-(3-i) [4i-3+i] [5i-3] 26. (3+2i)(2-3i) [6+4i-9i-6i^2] [6-5i+6] [12-5i] 27. 30+15i / 2-i [30+15i / 2-i * 2+i / 2+i] [(60+30i+30i+15i^2) / (4+1)] [(60+60i-15) / (5)] [(45+60i) / (5)] [9+12i] 28. 10-70i-3i+21i^2 / 1+49i^2 [(10-73i-21) / (1+49)] [(-11-73i) / (50)] [-11/50 - 73/50 * i] 29. 4i+(3-i) / 3-i [4i+(3-i) / 3-i * 3+i / 3+i] [(12i+4i^2+9-3i) / (9-i^2)] [(12i-4+9-3i) / (9+1)] [(5i+5) / (10)] [1/2 + 1/2 * i] 30. i^64 [(i^4)^16=1^16=1] 32. i^103 [(i^23)^4 * i^3] [1 * i^3=i^2 * i=-i] Solve by Completing the Square. 33. y^2-12y=4 [y^2-12y+36=4+36] [(y-6)^2=40] [y-6=±√40] [y=6±2√10] 34. x^2-4x+2=0 [x^2-4x+4=-2+4] [(x-2)^2=2] [x-2=±√2] [x=2±√2] 35. 3n^2+12n+10=0 [3(n+2)^2=2] [(n+2)^2=2/3] [n+2=±√2/3] [n=-2±√6/3] 36. 2n^2-3n-8=0 [n^2-3/2n-4=0] [(n-3/4)^2=23/16] [n-3/4=±√23/4] [n=3/4±√23/4] Consider the quadratic expression: y=x 2 +8x+6 Steps to complete the square: Move the constant term to the right side of the equation. y=x 2 +8x We complete the square by taking half of the coefficient of our x term, squaring it, and adding it to both sides of the equation. The coefficient of our x term is 8 , so half of it would be 4 , and squaring it gives us 16 . Because we’re adding the 16 inside the parentheses on the right where it’s being multiplied by 1 , we need to add 16 to the left side to make sure we’re adding the same thing to both sides.   y=x 2 +8x+16 Now we can rewrite the left side of the equation as a squared term. y=(x+4) 2 Vertex Form: The equation is now in vertex form: y=(x+4) 2 −10 Moving the constant term to the left side of the equation does not change the equation because adding zero to both sides does not change the equation.