Can you help with this x2 - 4x + 7 = 0 ?

Question

Consider only the discriminant, b^2 - 4ac, to determine whether one real-number solution, two different real-number solutions, or two different imaginary-number solutions exist.

Answer 1

x2-4x=-7

x2-x= -7/4

x= -7/4

x=-1.75

Answer 2

ok

so a=1   b=-4   c=7

16-4(1)(7)=16-28= -12 < 0 therefore : so x can accept no number and the question has got no result

Answer 3

3.5=x...

Answer 4

2x-4x+7=0

-2x+7=0

+2x      +2x

7=2x

okay i solved the first thing and I got 3.5.... I hope that helped

Answer 5

the b^2 -4ac results after substitution of the values  in -12

this indicates that there are imaginary solutions to the equation....using the quadratic formula  x = 2 +/-  i(SQRT3)

two imaginary solutions

Answer 6

we know that the discriminant in the quadratic formula is the number underneath the square root sign. the square root of a negative number is imaginary/unreal. and the square root of zero is zero so if you add/subtract it, nothing changes. if the discriminant is positive there are 2 different real number solutions. if it is negative then there are 2 different imaginary number solutions. and if it is 0, there is one real number solution.

in this case b² - 4ac =

(-4)² - 4(1)(7)

16 - 28

-12

it is negative. so 2 imaginary solutions.

Answer 7

I think it was "x^2 - 4x + 7 = 0" rite? Thennnn b^2 - 4ac = -12 ==> this equation will have two imaginary-number solutions. You wanna calculate the final exact answers? Get a calculator man!!

Anyway if I misunderstood any part of your question, inform me so that I can help out.  

Answer 8

a=1

b = -4

c = 7

discriminant = (-4)^2 - 4(1)7

= 16 - 28

= -12

Recall that we need to take the square root of the discriminant, and root(-12) yields the complex solution 2 + root(3)i

So in this case, the answer is two different complex (not imaginary numbered as the question calls them.) solutions

If the discriminant = 0, then there is one real number solution (Although, to be perfectly correct, it is two identical real number solutions)

If the discriminant is >0 , then there are two real number solutions.

Edit: There are a lot of wrong answers here.

People, if you don't understand the question, don't answer it.

Answer 9

let discriminant be D, if

If D > 0 then 2 different real solutions exist

if D = 0 then one real solution exists

if D < 0 then two imaginary solutions exist

Here D = 16 - 28 which is < 0

So imaginary solutions exist.

Answer 10

Do you mean -4x^2?

Answer 11

x^2 - 4x + 7 = 0

If you plug in the numbers for the discriminant you get (-4)^2 - (4)(1)(7) or -12.

Now... This is important because the discriminant is the part under the square root in the quadratic equation. Basically, if b^2 - 4ac is positive, it has two real solutions. If it's negative, you have two imaginary solutions. And if it's zero, then you only have one real solution.

Answer 12

Two different imaginary solutions exist.  I forgot the rule of how to do it with just the discriminant, so I worked out the whole thing using the quadratic formula.

Answers were:  2 - i(times radical3)   and   2 + i(times radical3)

Answer 13

a= 1

b= 4

c= 7

b^2 - 4ac = 4^2 - 4 x 1 x 7 = 16 - 28 = - 12

then, taking acount X= -b (+)(-) square root (b^2 - 4ac) / 2a

the solution is the third one. There are two different imaginay-number solutions.

it is square root (-12) = square root (12 i)

being i= square root (-1)

one solution: x = 4 + square root (12) i  / 2

another one: x = 4 - square root (12) i  / 2

Answer 14

x=3.5

Answer 15

b^2 - 4ac

Solving it we get ..

16-28 = -12 .

So..it will have imaginary roots (Tw0).