Use the quadratic formula to solve the equation. Give exact answers:
2x^2 -1 = 6x.
The choices are:
a) -3 + square root(7)/2, -3 - square root(7)/2
b) 3 + square root(11)/2, 3 - square root(11)/2
c) 3 + square root(7)/2, 3 - square root(7)/2
d) -3 + square root(11)/2, -3 - square root(11)/2
I attempted this problem but got none of these answers! Could you please walk me through the problem and tell me which answer is correct? Thank you so much! :)
2 x^2 - 6 x - 1 = 0
x = [ 6 +/- sqrt (36 + 8) ] / 4
x = [ 6 + /- sqrt (44) ] / 4
x = [ 6 + /- 2 sqrt (11) ] / 4
x = [ 3 +/- sqrt(11) ] / 2 so b
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2x^2-1=6x
2x^2-6x-1=0
a b c and then solve by using the quadratic formula
2009??? Everyone here must be in their 20s-30s. Nothing's changed much from then tho <3
Hello future students that might find this from years later! (I doubt anyone would see this but) my insta is @jaineykeyes if you ever want to know what I was like or what I do at the time of you seeing this! (I'm stuck and obsessed with the past if it isn't obvious enough)
To solve the equation 2x^2 - 1 = 6x using the quadratic formula, you need to rearrange the equation into the standard form of a quadratic equation: ax^2 + bx + c = 0.
In this case, your equation is already in the standard form. Rewrite it as follows:
2x^2 - 6x - 1 = 0
Now, you can identify the values of 'a', 'b', and 'c' for the quadratic formula:
a = 2
b = -6
c = -1
The quadratic formula is given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
Plug in the values of 'a', 'b', and 'c' into the formula and simplify:
x = (-(-6) ± √((-6)^2 - 4(2)(-1))) / (2(2))
x = (6 ± √(36 + 8)) / 4
x = (6 ± √44) / 4
x = (6 ± 2√11) / 4
x = (3 ± √11) / 2
Now, compare the solutions with the options provided:
a) -3 + √7/2, -3 - √7/2
b) 3 + √11/2, 3 - √11/2
c) 3 + √7/2, 3 - √7/2
d) -3 + √11/2, -3 - √11/2
Looking at the solutions obtained, we can identify that option b) matches the solutions we found:
3 + √11/2, 3 - √11/2
Therefore, the correct answer is b) 3 + √11/2, 3 - √11/2.
It's important to note that most quadratic equations will have two solutions, as indicated by the ± symbol in the quadratic formula.