Can a quadratic equation considered a quadratic function without a constant term

Mathematics

2 answers:

MArishka [77]3 years ago

6 0

Yes, its quadratic. constant may be zero.

Leviafan [203]3 years ago

3 0

Yes and zero will be the constant

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Numbers that add to -0.5 and multiply to 1

Oliga [24]

Short Answer
There are two numbers
x1 = -0.25 + 0.9682i <<<< answer 1
x2 = - 0.25 - 0.9582i <<<< answer 2

I take it there are two such numbers.
Let one number = x
Let one number = y

x + y = -0.5
y = - 0.5 - x (1)
xy = 1 (2)

Put equation 1 into equation 2
xy = 1
x(-0.5 - x) = 1
-0.5x - x^2 = 1 Subtract 1 from both sides.
-0.5x - x^2 - 1 = 0 Order these by powers
-x^2 - 0.5x -1 = 0 Multiply though by - 1
x^2 + 0.5x + 1 = 0 Use the quadratic formula to solve this.

$\text{x = }\dfrac{ -b \pm \sqrt{b^{2} - 4ac } }{2a}$

a = 1
b = 0.5
c = 1

$\text{x = }\dfrac{ -0.5 \pm \sqrt{0.5^{2} - 4*1*1 } }{2*1}$

x = [-0.5 +/- sqrt(0.25 - 4)] / 2
x = [-0.5 +/- sqrt(-3.75)] / 2
x = [-0.25 +/- 0.9682i

x1 = -0.25 + 0.9682 i
x2 = -0.25 - 0.9682 i

These two are conjugates. They will add as x1 + x2 = -0.25 - 0.25 = - 0.50.

The complex parts cancel out. Getting them to multiply to 1 will be a little more difficult. I'll do that under the check.

Check
(-0.25 - 0.9682i)(-0.25 + 0.9682i)
Use FOIL
F:-0.25 * -0.25 = 0.0625
O: -0.25*0.9682i
I: +0.25*0.9682i
L: -0.9682i*0.9682i = - 0.9375 i^2 = 0.9375

Notice
The two middle terms (labled "O" and "I" ) cancel out. They are of opposite signs.

The final result is 0.9375 and 0.0625 add up to 1

7 0

3 years ago

Use the Remainder Theorem to find the remainder for (2x^3-3x^2+6)/(x-1) and state whether or not the binomial is a factor of the

Bezzdna [24]

Answer:

Remainder= 5, and the binomial $(x-1)$ is not a factor of the given polynomial.