Consider only the discriminant, b2 - 4ac, to determine whether one real-number solution,?

Question

Consider only the discriminant, b2 - 4ac, to determine whether one real-number solution, two different real-number solutions, or two different imaginary-number solutions exist.

16x2 + 8x + 1 = 0

Answer 1

b^2 - 4ac =  8^2 - 4(16)(1)

              =  64 - 64

              =  0

Therefore, b^2 - 4ac = 0. This equation have one real-number solution.

Answer 2

When u have a quadratic equation (an equation where the highest power of x is 2, like u have here), then you can use what is called the Quadratic Formula to find out the values of x, and thus find the roots.

The formula is:

x = -b ± √(b^2 - 4ac)

               2a

The discriminant is the bit under the square root bracket. If the discriminant is less than zero, then u will have a number under the square root sign that is less than zero. As u know, u cannot take the square root of a negative number.

Another thing to note, is that a quadratic equation has EITHER two real solutions or two imaginary solutions. There is NO quadratic equation which has one real solution and one imaginary solution.

So having said all that, if the discriminant b^2 - 4ac is zero or greater than zero, then you will have two real solutions. If the discriminant is less than zero, then you will have two imaginary solutions.

In your question, a=16, b=8 and c=1 because u compare it with the formula ax^2 + bx + c.

So put a, b and c into the discriminant, and an answer zero or greater means two real solutions, and an answer less than zero means two imaginary solutions.

(PS ^ means square root, so x^2 means 'x to the power of 2' or 'x squared').

Answer 3

the discriminant (D) rule is -

a) D greater than zero but not a perfect sqaure :- two different real soln.

b) D greater than zero and a perfect sqaure , or equal to zero :- one real soln.

c) D is less than zero :- iaginary no. soln.

here in this equation-

D = 8^2 - 4X16X1 = 64-64 = 0

D is equal to zero.. means solution will be only one real no... i.e.

x= -1/4

Answer 4

You're looking at the stuff under the radical in the quadratic formula.

a (the coefficient before the squared term) = 16

b (coeff. before the x variable) = 8

c (the number with out a variable) = 1

plug them in:

(8)^2 - 4(16)(1) = 64- (64) = 0

therefore there is only one solution since you would be +/- zero to (b^2) /2

Answer 5

b^2-4ac in this case = 0.

zero is a real number so you have a one real-number solution.

y = -b +/- (sqrt of b^2-4ac) / 2a

so y = -b/2a = -1/4

Answer 6

The discriminant of the given equation=

8^2-4(16)(1)=0=>

the equation has 2 repeated real solutions,not one solution

because any polynomial of degree n should have n solutions.

Answer 7

This equation is of form ax^2+bx+c

a = 16 b = 8 c = 1

x=[-b+/-sqrt(b^2-4ac)]/2a]

x=[-8 +/-sqrt(8^2-4(16)(1)]/(2)(16)

discriminant is b^2-4ac =0

0 means one-real number solution (multiple root)

--------------------------------------...

x=[-8 +√(0)] / (2)(16)

x=[-8 -√(0)] / (2)(16)

x=[-8+0] / 32

x=[-8-0] / 32

The roots are -0.25 and -0.25