QUADRATIC EQUATIONS-help!?

Question

I need help solving these...

PLEASE TELL ME HOW

A) r² + 10r - 9 = 0

B) (2x +1)(x+3)=0

C) x²=4x+32

D) 2t² - 3T - 2 = 0

E) x²= 16

F) n² + 5n - 1 = 0

just teaching me how to solve any one of these would help.

thank you!!

Answer 1

D

2t² - 3t - 2 = 0

2t² - 4t + t- 2= 0

2t(t-2)+ 1(t-2) = 0

(2t+1) (t-2)= 0

either 2t+1=0 or t-2=0

therefore t=2 or t =-1/2

Answer 2

A is an easy one.

r² + 10r - 9 = 0

just name each one a, b, and c.   1= a, 10=b, and -9=c.

all you do is order it r²,r,and the number by itself.

the equation is (-b±√(b^2-4ac))/2a

(-10±√(10^2-4(1)(-9))/2(1)

(-10±√(100-4(-9))/2

(-10±√(100+36))/2

(-10±√(136))/2

the answer is in decimal.  I hope I showed enough for you to solve the rest of it.

Answer 3

A) r^2 + (10x-1)r -9

(r^2 + 10r) +(-r-9)

r(r+10) - (r + 9)

(r-1)(r+9)=0

r-1=0  r+9=0

r=1    r= -9

Answer 4

There are two ways to solve quadratic equations.  The first is to factor them, then set each of the factors equal to zero and solve for the variable.  The second is to use the quadratic formula.

Sometimes (like in A) you can't factor and so you have to use the equation.

B is already factored for you, so 2x+1=0, and sovle for x gives you -1/2, and x+3=0 solve for x gives you -3.  So, x = -1/2,3

C you need to set = to zero before you can solve, so move the 4x and 32 terms from one side to the other giving you x^2-4x-32=0.  When we factor that we get (x-8)(x+4)=0, so solving for x gives us x=8,-4

Solve D the same way.

E is a little different, when you set = to 0 you get x^2-16=0 so the middle term has dropped out.  That can only happen if you factor to (x-4)(x+4)=0 so that when you multiply out using FOIL (first, outside, inside, last) the middle terms cancel to zero.  That will give you x=4, -4

See if you can figure out F.