can anyone please help me with some algebra problems?
so if you can work them out for me.
thanx !!!
1. solve (x-2)SQUARED=25
2. solve 5xSQUARED + 30x = 0
3. what are the x-intercepts of the equation 5xSQUARED – 20 = 0
the SQUARED thingies are like.
the the second power.
Favorite Answer
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1. open up = (x-2) * (x-2) = 25
x squared – 2x -2x +4 = 25
x squared -4x +4 =25
x squared -4x +4 -4 =25-4
x squared – 4x = 21
factor out x
x(x-4) =21
………..im stuck ……get someone to help you from there im sure im on the right track just that i dont remember if you get the log on both sides or something else……. 🙂
2. in the second question take 30x to the other side
5xsquared = -30x
divide both sides by x
5x squared/x = -30x/x
5x= -30
divide both sides by 5
5x/5 = -30/5
x = – 6 (negative 6)
3. 5x squared -20 = 0
take 20 on the other side
5x squared – 20 +20 = 0 +20
5x squared = 20
divide both sides by 5
5 x squared / 5 = 20/5
x squared = 4
get square root
x = 2 the other intercept is x= – 2
(negative x negative = positive) (-2 x -2 = +4)
Have fun with the math
so (x-2)(x-2) = 25
now, you need to work further:
(x-2)(x-2) = 25
x^2 – 4x + 4 = 25
x^2 – 4x = 25 – 4
x(x – 4) = 21
x = 7
2) 5x^2 + 30x = 0
(x-2)squarroot = (25)squareroot
(x-2) = 5
x-2 = 5
adding 2 0n both sides
x-2+2 = 5+2
x+0 = 7
x = 7 solved
2. taking x common from equation
x(5square x + 30) = 0
5square x +30 =0
25x + 30 = 0
25x = -30
x = 30/25 dividing both sides by 25
so x = 6/5 solved
x – 2 = squareroot of 25
x = 5
2.) 5x squared + 30x = 0
5x squared = -30x
5x squared divided by x = -30
5x = -30
x = -30 divided by 5
x = -6
(x – 2) (x – 2) = 25
Simply you could say x = 7 but x = -3 too
remove the brackets you get
x^2 – 4x + 4 = 25 therefore
x^2 – 4x – 21 = 0 you go from here.
5x^2 + 30x = 0
5(x^2 + 6x) = 0
x can = +6 or -6
follow the same steps for three.
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