Question

please solve this. ..........

i need all 3 please​

Answer 1

Answer:

3 ) Notice that the equation involves three powers of x ? There’s x2 , x1(=x) and x0=1 . This means that we are dealing with a quadratic equation in terms of x . This, in turn, means that the equation will have two roots, not one - so your use of ‘the root’ is inappropriate. Of course, these roots might be identical; let’s check.

For the general quadratic ax2+bx+c=0 , we know that the roots are given by the formula:

x=−b±b2−4ac√2a

From this, it should be obvious that the two roots are identical when the expression under the radical sign, which we call the discriminant, is equal to zero. Is this the case for our equation?

b=1⇒b2=1

a=1 and c=1⇒4ac=4

∴b2−4ac=1−4=−3≠0

So, we are looking at two distinct roots; further, as the discriminant is negative and the quadratic has ‘real’ coefficients, the two roots form a complex conjugate pair.

Let’s use the formula to find these roots:

x=−1±−3√2=−1±i3√2

[math]= -\frac {1}{2} \pm \frac {\sqrt{3}{2}i[/math]

We can rewrite this as:

x1=cos(23π+2nπ)+isin(23π+2nπ)

x2=cos(43π+2nπ)+isin(43π+2nπ)

where n is any integer

Of course, there’s another way. To demonstrate this, let’s multiply your quadratic expression by (x−1) .

(x2+x+1)(x−1)=0

⇒(x2+x+1)x−(x2+x+1)=0

⇒(x3+x2+x)−(x2+x+1)=0

⇒x3−1=0

As a cubic equation, this has three roots, which are the three cubic roots of 1. The term we multiplied the original quadratic by, namely (x−1) , gives us the ‘real’ root of x=1 , hence the roots of the original equation must give us the other cube roots of 1.

Step-by-step explanation:

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Answer 2

Answer:

2. Discriminant =

$( - 4) {}^{2} - 4 \times 4 \times 1 = 16 - 16 = 0$

so roots are Real & Equal.