ax2+ bx+c=0

Where a, b, and c are integers

and a≥1

I. To solve an equation in the form ax2+c=k, for some value k. This is the simplest quadratic equation to solve, because the middle term is missing. Strategy: To isolate the square term and then take the square root of both sides. Ex. 1) Isolate the square term, divide both sides by 2

Take the square root of both sides

2x2=40

2x22= 40 2

x2 =20

Remember there are two possible solutions

x2= 20

Simplify radical; Solutions

x= ± 20

x=± 25

(Please refer to previous instructional materials Simplifying Radical Expressions )

II. To solve a quadratic equation arranged in the form ax2+ bx=0. Strategy: To factor the binomial using the greatest common factor (GCF), set the monomial factor and the binomial factor equal to zero, and solve. Ex. 2) 12x2- 18x=0

6x2x-3= 0Factor using the GCF

6x=0 2x-3=0Set the monomial and binomial equal to zero

x=0 x= 32Solutions

* In some cases, the GCF is simply the variable with coefficient of 1. III. To solve an equation in the form ax2+ bx+c=0, where the trinomial is a perfect square. This too is a simple quadratic equation to solve, because it factors into the form m2=0, for some binomial m. (For factoring instructional methods, select The Easy Way to Factor Trinomials ) Strategy: To factor the trinomial, set each binomial equal to zero, and solve. Ex. 3) x2+ 6x+9=0

x+32=0Factor as a perfect square

x+3x+3= 0Not necessary, but valuable step to show two solutions x+3=0 x+3=0Set each binomial equal to zero

x= -3 x= -3Solve

x= -3Double root solution

IV. To solve an equation in the form ax2+ bx+c=0, where the trinomial is not a perfect square, but factorable. Similar to the last example, this is a simple quadratic equation to solve, because it factors into the form mn=0, for some binomials m and n. Strategy: To factor the trinomial, set each binomial equal to zero, and solve. Ex. 4) 2x2-x-6=0

* Using the factoring method from The Easy Way to Factor Trinomials, we need to find two number that multiply to give ac, or -12, and add to give b, or -1. These values are -4 and 3. Rewrite the trinomial with these two values as coefficients to x that add to the current middle term of -1x.

2x2- 4x+3x-6=0Rewrite middle term

2x2- 4x+3x-6=0

2xx-2+ 3x-2= 0Factor by grouping

x-22x+3= 0Factor out the common binomial (x-2)

x-2=0 2x+3=0Set each binomial equal to zero

x=2 x= -32Solutions

V. To solve a quadratic equation not arranged in the form ax2+ bx+c=0, but factorable.

Strategy: To combine like terms to one side, set equal to zero, factor the trinomial, set each

binomial equal to zero, and solve.

Ex. 5) 6x2+ 2x-3=9x+2

-9x -9x

6x2- 7x-3= 2

-2 -2

6x2- 7x-5=0

* To factor this trinomial, we are looking for two numbers that multiply to give ac, or -30, and add to give b, or -7. These values would be 3 and -10. Rewrite the trinomial with these two values as coefficients to x that add to the current middle...