a.
6x2 = 36
x2 = 36/6
x2 = 6
x = sqrt(6)
b. and c. can be solved similarly.
Post if you could use additional help or would like to check answers.
Please Solve:
a. 6x^2 = 36
b. 2x^2-20= 0
c. 4x^2+12=0
6 answers
Could it be root - 6 as well cause
-6*-6 = 36?
-6*-6 = 36?
±sqrt(6) would be correct, approximately equal to ±2.449489742783178...
( 2.449...)2 = 6
(-2.449...)2 = 6
But sqrt(-6) would equal to sqrt(6)*i, a complex number.
( ±2.449...*i) 2
= (±2.449...)2 * i 2
= 6*(-1)
=-6
( 2.449...)2 = 6
(-2.449...)2 = 6
But sqrt(-6) would equal to sqrt(6)*i, a complex number.
( ±2.449...*i) 2
= (±2.449...)2 * i 2
= 6*(-1)
=-6
So am i correct? Can i use both root 6 and -root 6
The two roots are indeed root 6 and -root 6, but not root(-6).
Got it thanks!
8 answers