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Quadratic Equat.1

Solving equations by the quadratic formula and quadratic methods

By knowing this value, we can discriminate among the possible number and type of solutions of a quadratic equation. Discriminant
What are the steps to solving a quadratic equation? 1. If the equation is in the form (ax+b)^2, use the square root property and solve. If not, write the equation in standard form (ax^2+bx+c=0) 2.Try to solve the equation by the factoring method. If not possible, use quadratic formula
Solve p^4-3p^2-4=0 (p^2-4)(p^2+1)=0 (p-2)(p+2)(p^2+1)=0 p-2=0 or p+2=0 or p^2+1=0 p=2; p=-2; p=+-i (the square root of -1)
Define quadratic equation. A second degree equation which is written as Ax^2+Bx+C=0
Solve this equation: m^2-m-2=0 m^2-m-2=0 (m-2)(m+1)=0 m-2=0 or m+1=0 m=2, m=-1
Use the quadratic formula to solve this equation: x^2+7x+4=0 -b=-7; (b^2-4ac)=(7^2-4(1)(4); 2a=(2)(1) (-7+-√49-16)/2 (-7+-√33)/2 Solutions: (-7+√33)/2 and (-7-√33)/2
Use the quadratic formula to solve this equation: x^2+3x=1 x^2+3x-1=0 (-(3)+-√(3)^2-(4)(1)(-1))/(2)(1) (-3+-√9-(-4))/2 Solutions: (-3+√13)/2 and (-3-√13)/2
Use the quadratic formula to solve this equation: (x+5)(x=3)=5 Multiply the binomials: x^2+8x+15-5=5-5 x^2+8x+10=0 (-(8)+-√(8)^2-4(1)(10))/(2)(1) (-8+-√64-40)/2 (-8+-√24)/2=(-8+-2√6)/2 (-8+2√6)/2 or (-8-2√6)/2 Solutions: -4+√6 and -4-√6
If b^2-4ac is positive, There will be two real solutions.
If b^2-4ac is negative, There will be two complex solutions, but they will not be real.
If b^2-4ac is zero, There will be one real solution.
Created by: kaylaroszkowski