A) Completing squares we got the solutions:
x = -1/4 ± √(33/16)
B) Using the quadratic formula method we got:
x = 1 ± √7
How to complete squares?
Remember the relation:
(a + b)^2 = a^2 + 2ab + b^2
Now, we have the quadratic equation:
2x^2 + x - 4 = 0
First, we can rewrite our equation as:
(√2*x)^2 + 2*(1/2)*x - 4 = 0
Now, we can add and subtract (1/√2*2)^2 so we get:
(√2*x)^2 + 2*(1/2)*x - 4 + (1/√2*2)^2 - (1/√2*2)^2 =0
[(√2*x)^2 + 2*(1/√2*2)*(√2*x) + (1/√2*2)^2] - 4 - (1/√2*2)^2 = 0
Now we can complete squares inside the first part:
(√2*x + (1/√2*2))^2 - 4 - (1/√2*2)^2 = 0
(√2*x + (1/√2*2))^2 - 4 - 1/8 = 0
(√2*x + (1/√2*2))^2 = 4 + 1/8 = 33/8
Now we can solve this for x:
(√2*x + (1/√2*2)) = ±√(33/8)
√2*x = - (1/√2*2) ± √(33/8)
x = ( -(1/√2*2) ± √(33/8))/√2
x = -1/4 ± √(33/16)
These are the two solutions of x.
How to use the quadratic formula method?
For a general quadratic equation:
a*x^2 + b*x + c = 0
The solutions are given by:
x = (-b ± √(b^2 - 4ac))/2a
In this case, we have:
x^2 - 2x - 6 = 0
Then the solutions are:
x = (2 ± √( (-2)^2 - 4*1*(-6))/2*1
x = (2 ± √(4 + 24))/2
x = 1 ± √(28)/2
x = 1 ± √7
If you want to learn more about quadratic equations
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