How do I find the root of x(sqared)+ x +1 = 0?
the two roots are given by r1 &r2 = [-b+-(b^2-4ac)^1/2]/2a where a=1 b=1 c=1 (according to the given eqn)
What are the roots of (x-1) ^3+8 =0?
If we call any cube root of unity [math]w[/math], so that [math]w^3 = 1,[/math] we can write this as[math](x-1)^3 = -8w^3[/math]Taking cube roots,[math]x-1 = -2 w[/math][math]x = 1 - 2w[/math]There are three [math]w[/math]s so there are three solutions:[math]w^3 - 1 = 0[/math][math](w-1)(w^2 + w + 1) = 0[/math][math]w = 1[/math] or [math]w = \dfrac{-1 \pm i \sqrt{3}}{2}[/math][math]x = 1 -2w[/math] so [math]x=-1[/math] or [math]x=2 \pm i\sqrt{3}[/math]Check: Obviously [math]x=-1[/math] works. We’ll check one of the others:[math](2-i\sqrt{3} - 1)^3 +8 = (1-i\sqrt{3})(1-i\sqrt{3})^2+8=(1-i\sqrt{3})(-2 -2i\sqrt{3})+8=-8+8 \quad\checkmark[/math]
How do I find all roots of [math]x^2-3x-\sqrt{x^2-3x}-2 = 0[/math]?
Let y^2 = x^2 - 3xy^2 - y - 2 = 0y^2 - 2y + y - 2 = 0y(y - 2) + 1 (y - 2) = 0(y+1)(y-2) = 0y = -1 or y = 2ThenBy resubstitution -x^2 - 3x = -1x^2 - 3x + 1 = 0By shreedharacharya formula -x = {3 ± √(9–4)}/4x = ( 3 ± √5 ) / 4Also -x^2 - 3x = 2x^2 - 3x - 2 = 0By shreedharacharya formula -x = {3 ± √(9+16)}/4x = ( 3 ± 5 ) / 4x = 2 or x = -1/2x = 2, -1/2, (3+√5)/4, (3-√5)/4A2A. THANK-YOU!
How do I find the roots of these equations: x^3- 7X^2+11x- 5=0.....X^3+3X^2+3X=1=0....X^ 3+5X^2-8X-48=0?
x^3- 7X^2+11x- 5=0 or (x-1)^2(x-5) roots x=1 double root or x=5 X^3+3X^2+3X=1=0 or (x+1)^3 =0 x = -1 triple root X^ 3+5X^2-8X-48=0 or (x+4)^2(x-3)=0 x= -4 double root x=3
Find remaining roots. 2x^3-5x^2-4x+3=0. root, x=3?
Given: 2x^3-5x^2-4x+3=0 Since we're told that one root is 3, we can divide 2x^3-5x^2-4x+3 by (x-3) to get: 2x^2 + x - 1 = 0 This can then be factored to: (2x-1)(x+1) = 0 Thus: 2x-1=0 2x = 1 x = 1/2 OR x+1 = 0 x = -1 So the three roots are: x = 3 (given) x = 1/2 x = -1 .
Find only the rational roots of 4x3 – 3x – 1 = 0. need help with these algebra questions!?
1. Find only the rational roots of 4x3 – 3x – 1 = 0. 2. Find the roots of 2x3 + 8x2 + 7x + 3 = 0. 3. The possible rational roots of x3 + 3x2 – 4x + 2 are +/- 3 and +/- 2. true or false?
What is the root of x^2+x+1=0?
What is the root of x^2+x+1=0?Notice that the equation involves three powers of [math]x[/math]? There’s [math]x^2[/math], [math]x^1 (= x)[/math] and [math]x^0 = 1[/math]. This means that we are dealing with a quadratic equation in terms of [math]x[/math]. 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 [math]ax^2 + bx + c = 0[/math], we know that the roots are given by the formula:[math]x = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a}[/math]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?[math]b = 1 \Rightarrow b^2 = 1[/math][math]a = 1[/math] and [math]c = 1 \Rightarrow 4ac = 4[/math][math]\therefore b^2 - 4ac = 1 - 4 = -3 \neq 0[/math]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:[math]x = \frac {-1 \pm \sqrt{-3}}{2} = \frac {-1 \pm i\sqrt{3}}{2}[/math][math]= -\frac {1}{2} \pm \frac {\sqrt{3}{2}i[/math]We can rewrite this as:[math]x_1 = \cos(\frac {2}{3}\pi + 2n\pi) + i\sin(\frac {2}{3}\pi + 2n\pi)[/math][math]x_2 = \cos(\frac {4}{3}\pi + 2n\pi) + i\sin(\frac {4}{3}\pi + 2n\pi)[/math]where [math]n[/math] is any integer[math]\\[/math]Of course, there’s another way. To demonstrate this, let’s multiply your quadratic expression by [math](x - 1)[/math].[math](x^2 + x + 1)(x - 1) = 0[/math][math]\Rightarrow (x^2 + x + 1)x - (x^2 + x + 1) = 0[/math][math]\Rightarrow (x^3 + x^2 + x) - (x^2 + x + 1) = 0[/math][math]\Rightarrow x^3 - 1 = 0[/math]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 [math](x - 1)[/math], gives us the ‘real’ root of [math]x = 1[/math], hence the roots of the original equation must give us the other cube roots of 1.Plotting these roots on an Argand diagram, they all lie on a unit circle (shown in blue) and are separated by an equal angle (120° aka ⅔π radians).
If a,b, c are the roots of x^3-x^2-1=0. Then what will be the value of Σ ((1+a) /(1-a))?
Answer will be -5.since a , b , c are the roots of x^3-x^2–1. Then as per the the property of polynomials.Sum of roots taken 1 at a time =-b/aSum of roots taken 2 at a time =c/aSum of roots Taken 3 at a time=-d/a...And the sequence continuous for all higher order equations.Since given equation is a cubic equation.thenEquation 1 : a+b+c=1Equation 2: ab+bc+ac=0Equation 3 : abc=1Replace 1+a in Σ ((1+a) /(1-a)) with 2-(1-a) and use above 3 equation and simplify Σ ((1+a) /(1-a)).You will get the answer as -5.Hope above hints helps.