SOLUTION: I need some help. It simply says find the solutions but I am not sure what method it is asking me to solve it with. I don't know if it means solve using the quadratic equation or
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-> SOLUTION: I need some help. It simply says find the solutions but I am not sure what method it is asking me to solve it with. I don't know if it means solve using the quadratic equation or
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Question 559425: I need some help. It simply says find the solutions but I am not sure what method it is asking me to solve it with. I don't know if it means solve using the quadratic equation or if there is some other way I am suppose to be doing it. Please help! Thanks.
Find the solutions:
4) (m+3)^2 + 5(m+3) –14 = 0
5) (2x-1)^2 –4(2x-1) + 2 = 0
You can put this solution on YOUR website! these are quadratic equations.
instead of x, you have (m+3)
instead of x, you have (2x-1)
assign the expression inside the parentheses to a variable other than a variable that is being used in the parentheses and you'll see that more clearly.
for your first problem:
(m+3)^2 + 5(m+3) - 14 = 0
let x = m+3
your equation becomes:
x^2 + 5x - 14 = 0
this factors to:
(x+7) * (x-2) = 0
x = -7 or x = 2
for your second problem:
(2x-1)^2 - 4(2x-1) + 2 = 0
let y = 2x-1
your equation becomes:
y^2 - 4y + 2 = 0
this doesn't factor easily.
use the quadratic formula on this.
y = (-b +/- sqrt(b^2-4ac))/(2a).
a = 1
b = -4
c = 2
-----
-b = 4
sqrt(b^-4ac) = sqrt(8)
2a = 2
-----
y = (4 + sqrt(8))/2 = 3.414213562
y = (4 - sqrt(8))/2 = 0.585786438
------
since we set:
y = 2x-1, then we get:
2x-1 = 3.414213562 which results in:
x = 2.207106781
and we get:
2x-1 = .585786438 which results in:
x = 0.792893219
there is no need to go back to the original equation of:
(2x-1)^2 - 4(2x-1) + 2 = 0 because when we calculate 2x-1, we get y.
since y is the value we used to solve the equation, we will get the same answer because:
x = 2.207106781 times 2 minus 1 will equal 3.414213562 which is the value of y.
similarly:
x = 0.792893219 times 2 minus 1 will equal .585786438 which is the value of y.
these 2 values of y are what we solved for originally.
You can put this solution on YOUR website! (m+3)^2 + 5(m+3) –14 = 0
using the quadratic eqaution
(m+3)^2+7(m+3)-2(m+3)-14=0
(m+3)(m+3+7)-2(m+3+7)=0
(m+3)(m+10)-2(m+10)=0
(m+10)(m+3-2)=0
(m+10)(m+1)=0
m=-10 OR -1
