Asked by jenna
please help me solve these
1. 2√5/8 + 4√3/8 - 1/2√68
2. √4k-5 +1=8
(the radical sign goes over 4k-5)
3. (x+4)*2 /3 =8
4. Write the value of the discriminant of each equation. Then use it to decide how many different real-number roots the equation has.
x*2-6x+2=0
1. 2√5/8 + 4√3/8 - 1/2√68
2. √4k-5 +1=8
(the radical sign goes over 4k-5)
3. (x+4)*2 /3 =8
4. Write the value of the discriminant of each equation. Then use it to decide how many different real-number roots the equation has.
x*2-6x+2=0
Answers
Answered by
L
Answer to number 3:
2/3(x+4) = 8
2(x+4) = 24
2x+8 = 24
2x = 16
x = 8
2/3(x+4) = 8
2(x+4) = 24
2x+8 = 24
2x = 16
x = 8
Answered by
L
Answer to number 3 is wrong. (didn't realise it was x+4 to the power of two thirds. Sorry):
Answered by
Reiny
Looks like Jenna is using * as the sign for exponents.
In #3 , no brackets were used around the 2/3, so it has to be read as
((x+4)^2)/3 = 8
then
(x+4)^2 = 24
x+4 = +/- sqrt(24)
x = -4 +/- 2sqrt(6)
In #3 , no brackets were used around the 2/3, so it has to be read as
((x+4)^2)/3 = 8
then
(x+4)^2 = 24
x+4 = +/- sqrt(24)
x = -4 +/- 2sqrt(6)
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