Question

The roots of equations x^ +x+1=0 are

Answer 1

Step-by-step explanation:

1.618 & -0.618 are the roots of quadratic equation x2 - x - 1 = 0.

Answer 2

Step-by-step explanation:

Solution of quadratic equation( ax2+bx+c=0) is given by -

x=−b±b2−4ac√2a . . . (1)

Here, a = 1, b = 1 and c = 1 (x2+x+1=0)

Just put the value of a, b and c in eq(1),

x=−1±12−4∗1∗1√2∗1

x=−1±−3√2

This is shows its have only imaginary roots.

As i=−1−−−√.

x=−1±i3√2

So two roots are,

x=−1+i3√2 and x=−1−i3√2.

Hope it helps you !

f(x)=x2+x+1

Roots of the equation f(x)=0 will be

x=−b±b2−4ac−−−−−−−√2a

x=−1±12−4(1)(1)−−−−−−−−−−√2(1)

x=−1±3–√i2

x=−1+3–√i2 and

x=−1−3–√i2

are also complex cube roots of 1 and also denoted respectively as ω and ω2 .

How do I factorise x²+x+1?

What is the root of x^2+x+1=0?

What is the answer for (x²-x+1) ²?

The given equation clearly does not have 1 as it's root.

Multiply both sides of the the given equation by (x−1) .

This gives you equation x3–1=0 .

1 is the only root of the given cubic. The other two are the complex cube roots of 1.

They are {[math]\[/math] cos(2π/3)+isin(2π/3),cos(4π/3)+isin(4π/3) }.

This can also be obtained by directly applying the quadratic formula.

Use the normal formula for solving quadratic equations-

x=(-1 +/- sqrt(1–4*1*1))/2*1

x=(-1 +/- sqrt(-3))/2 (square root of -3 is sqrt(3)*i as sqrt(-1*3)=sqrt(3)*sqrt(-1)=sqrt(3)*i

So the roots are

x=(-1/2) + sqrt(3)*i/2 and x=(-1/2) - sqrt(3)*i/2

Or

x=-0.5 + 0.866025404 i and x=-0.5 - 0.866025404 i

For a quadratic equation,

ax2+bx+c=0

Roots are given by( Shridharacharya method ):-

x=−b±b2−4ac√2a

Therefore, for this question, a=1,b=1,c=1

So, x=−1±1−4√2

⇒x=−1±i3√2

so, there are two complex roots of the given equation, and no real roots.

Just for information, the roots which are obtained are called complex cube roots of unity, because they are solutions of:-

x3−1=0

General form of an eqn is

ax²+bx+c=0

Given eqn is (1)x²+(1)x+1=0

This implies, a=1; b=1; c=1

Terefore, by Quadratic formula

x=[-b±√(b²-4ac)]/2a

x=[-1±√(1²-4×1×1)]/2×1

x=[-1±√(-3)]/2

x=[-1±√3×√-1]/2

But √-1=i (a complex no. called iota)

x=[-1±i√3]/2

x=[i√3–1]/2 and -[i√3+1]/2