Please Solve:
a. 6x^2 = 36
b. 2x^2-20= 0
c. 4x^2+12=0
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.
Could it be root - 6 as well cause
-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
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!
a. To solve the equation 6x^2 = 36, we need to isolate x. Start by dividing both sides of the equation by 6:
6x^2 / 6 = 36 / 6
x^2 = 6
To get rid of the exponent, we can take the square root of both sides:
√(x^2) = √6
x = ±√6
So the solution to the equation is x = ±√6.
b. To solve the equation 2x^2 - 20 = 0, we need to isolate x. Begin by adding 20 to both sides:
2x^2 - 20 + 20 = 0 + 20
2x^2 = 20
Next, divide both sides of the equation by 2:
(2x^2) / 2 = 20 / 2
x^2 = 10
To remove the exponent, take the square root of both sides:
√(x^2) = √10
x = ±√10
So the solution to the equation is x = ±√10.
c. To solve the equation 4x^2 + 12 = 0, we need to isolate x. Begin by subtracting 12 from both sides:
4x^2 + 12 - 12 = 0 - 12
4x^2 = -12
Next, divide both sides of the equation by 4:
(4x^2) / 4 = -12 / 4
x^2 = -3
Since we end up with a negative number under the square root, there are no real solutions in this case. The equation has no real solutions.
In summary, the solutions to the given equations are:
a. x = ±√6
b. x = ±√10
c. There are no real solutions.