5141 Shares

Topic: **Problem solving by quadratic equations****Question:**
I'm learning solving quadratic equations by factoring can you help me with sum of these problems please & thanks you :)
1. x²+4x-12 = 0
2. x²-4x=21
3. n²-4= -3n
4. 3x²+13x+4=0
5. 3p²+8p+5=0
6. 3r²+10r+9=6
7. 2n²+17n+11=3
Thanks

July 16, 2019 / By Ava

x^2+4x-12=0 factors of twelve that will give you a middle term of 4x are 6 and 2 so factors are (x+6)(x-2) pay attention to signs if you need to solve for x, then set bothe parenthesis to =0 and x=-6, 2 and that's how you should do the rest of the problems ))

👍 142 | 👎 7

Did you like the answer? We found more questions related to the topic: **Problem solving by quadratic equations**

You should have the following patterns so drilled into your head by countless homework exercises that you recognize them whether the numbers are constant or variables. a^-b^2 = (a-b)(a+b) (a+b)^2 = a^2 + 2ab + b^2 (a-b)^2 = a^2 -2ab + b^2 In high school this is the same as 3 time 7 equals 21 in third grade. If you plan to learn math without doing any F**** homework, now is the time to use the technique you planned to substitue.

1. (x+2)(x-6)=0 x+2=0 or x-6=0 (below) (subtract 2 from both sides) x = -2 or x-6=0 (add six to both sides) x = 6 so, x= -2 or 6 2. (x-2)(x-4)=0 x-2=0 or x-4=0 (same procedure as above) x=2 or x= 4

1. x²+4x-12 = 0 (x-2)(x+6)=0 2. x²-4x=21 0=x^2-4x-21 then, (x+3)(x-7)=0 3. n²-4= -3n 0=n^2+3n-4 then, (n-1)(n+4) 4. 3x²+13x+4=0 (3x+1)(x+4)=0 5. 3p²+8p+5=0 (3p+5)(p+1)=0 6. 3r²+10r+9=6 = 3r^2+10r+3 so, 0=(3r+1)(r+3) 7. 2n²+17n+11=3 =2n^2+17n+8 so, 0=(2n+1)(n+8)

👍 50 | 👎 3

Im only solving two because this could be your hw so 1. To factor 1st find 2 numbers that when multiplied equal -12. Then find a factor pair that when added equals 4 So the pair would be 6 and -2 The answer would be (x+6)(x-2) 2. First move the 21 to the other side which would make x^2-4x-21 Then find factor pairs that would be -7 and 3 Answer is (x-7)(x+3) for those that have a number in front of x^2 make sure the pairs that one of them is multiplied by the number in front

👍 44 | 👎 -1

2x^2 - x - 21 = 0 (multiply 2 and -21, which is -42; find two integers that multiply to -42 but add to -1, which are 6 and -7) 2x^2 + 6x - 7x - 21 = 0 (group terms into pairs) (2x^2 + 6x) + (-7x - 21) = 0 (factor each group) 2x(x + 3) - 7(x + 3) = 0 (make (2x - 7) a factor and (x + 3) a factor) (2x - 7)(x + 3) = 0 (zero-product property) 2x - 7 = 0 or x + 3 = 0 2x = 7 or x = -3 x = 7/2 or x = -3 <===ANSWER 3x^2 + 14x - 5 = 0 (multiply 3 and -5, which is -15; find two integers that multiply to -15 but add to 14, which are 15 and -1) 3x^2 - x + 15x - 5 = 0 (group into pairs) (3x^2 - x) + (15x - 5) = 0 (factor each group) (x(3x - 1) + 5(3x - 1) = 0 (make (x + 5) a factor and (3x - 1) a factor) (x + 5)(3x - 1) = 0 (zero-product property) x + 5 = 0 or 3x - 1 = 0 x = -5 or 3x = 1 x = -5 or x = 1/3 <===ANSWER 3x^2 + 11x - 4 = 0 (multiply 3 and -4, which is -12; find two integers that multiply to -12 but add to 11, which are -1 and 12) 3x^2 + 12x - x - 4 = 0 (group into pairs) (3x^2 + 12x) + (-x - 4) = 0 (factor each group) 3x(x + 4) - 1(x + 4) = 0 (make (3x - 1) a factor and (x + 4) a factor) (3x - 1)(x + 4) = 0 (zero-product property) 3x - 1 = 0 or x + 4 = 0 3x = 1 or x = -4 x = 1/3 or x = -4 <===ANSWER Factoring by this method is called the AC method: http://www.tcc.edu/vml/Mth03/Trinom/tri7... It works on factoring any quadratic (if it is factorable to begin with, of course).

If you have your own answer to the question problem solving by quadratic equations, then you can write your own version, using the form below for an extended answer.