Can someone please explain this ?

Question

For the quadratic equation 2x^2 – 4x = 1:
A. find the discriminant
B. According to the discriminant, how many solutions does the equation have?
C. According to the discriminant, are the solutions real or complex numbers?
D. Solve the equation using the quadratic formula.

Answer 1

Quadratic Formula

x = - b ± √b² - 4ac / 2a

2x² - 4x = 1

2x² - 4x - 1 = 1 - 1

2x² - 4x - 1 = 0

a = 2

b = - 4

x = - 1

x = - (- 4) ± √(-4)² - 4(2)(- 1) / 2(2)

x = 4 ± √16 - (- 8) / 4

x = 4 ± √16 + 8 / 4

x = 4 ± √24 / 4

x = 4 ± 4.898978486 / 4

- - - - - - - - - - -

Solving for +

x = 4 + 4.898979486 / 4

x = 8.898979486 / 4

x = 2.224744871

- - - - - - - -

Solving for -

x = 4 - 4.898979486 / 4

x = - 0.898979486 / 4

x = 0.224744871

- - - - - - - - -

A . The discriminate √24

B. Thee are two solutions

C. Real numbers

D. The equation has been solved.

- - - - - - -s-

Answer 2

When you're doing quadratics, the discriminant is that thing under the radical sign in the quadratic formula...
b^2 - 4ac is the discriminant (where a, b, and c are the coefficients on the x^2, x, and constant terms, respectively).

Remember, the quadratic looks like
ax^2 + bx + c = 0.

Your equation does not.

We need to subtract 1 from each side to put it in quadratic form.
2x^2 - 4x -1 = 0

The discrimanant is b^2 - 4ac, or (-4)^2 - 4(2)(-1) or
16 - -8
or 16 + 8 = 24
According to the discriminant, this has 2 solutions. If the discriminant were 0, it would have 1 solution, if the discriminant were < 0 (negative) it would have zero solutions.

So, x = -(-4) / (2*2) + sqrt(24)/(2*2) and x=-(-4)/(2*2) - sqrt(24)/(2*2)
or
1 + sqrt(24)/4
and
1 - sqrt(24)/4

Answer 3

A: The discriminant is D=b^2-4ac.

Where the equation is in the form y = ax^2 + bx +c

We can put your equation in this form by moving the one onto the the other side so:

y= 2x^2-4x-1.

Therefore D=(-4)^2 -4*2*(-1)=16+8=24

B:
When D > 0 , P(x) has two distinct real roots.
When D = 0, P(x) has two coincident real roots(two the same)
When D < 0 , P(x) has no real roots, and its graph lies strictly above or below the x-axis. In this case,

So we can see we will have two real roots.

C:Real

D; Quadratic formula is x= [-b(+or-)(b^2-4ac)^(1/2)]/2a

x=[-2(+or-)(24)^1/2]/4 =[ -2(+or-)2(6)^1/2]/4

So your two solutions are:
= -1/2+[(6)^1/2]/2 and -1/2 - [(6)^1/2]/2

Answer 4

If you have a quadratic in the form of ax² + bx + c = 0, then the discriminant is b² - 4ac. If the discriminant is less than zero, you have no real solutions. If it's equal to zero, it has one real solution. If it's greater than zero, you have two different real solutions. This makes sense if you look at where the disciminant shows up in the quadratic forumla.

Answer 5

A. 16+8
B. real
C. (4+(24)^(1/2))/4
=1+(3/2)^(1/2)
and 1-(3/2)^(1/2)

Answer 6

234