quadratic polynomials+equations

Analyzing the quadratic, we can see that $A + B = 4$ and $AB = 1$ .

Since $A^2 + B^2 = (A+B)^2 - 2AB$ and $A^3 + B^3 = (A^2 - AB + B^2)(A+B)$ , we can simply plug the values to find out that $A^2 + B^2 = 4^2 - 2 = 14$ and thus $A^3 + B^3 = (14 - 1) \times 4 = \boxed{52. \:}$

For a quadratic equation $ax^{2}+bx+c=0$ , sum of the roots is $\frac{-b}{a}$ and product of the roots is $\frac{c}{a}$ .

So, here A + B = 4 and A*B = 1.

We need to calculate $A^{3} + B^{3}$ :

$A^{3} + B^{3} = (A + B)(A^{2} + B^{2} - AB)$ ..... Identity of sum of cubes

= $(A + B)((A+B)^{2} - 2AB - AB)$ ..... Identity of sum of squares

Substitute the values of A + B and AB,

= (4)( $4^{2} - 2\times1 - 1)$

= $\boxed{52}$

That's the answer!

From the property of quadratic polynomials, it is clear that $A + B = 4$ and $AB = 1$

$(A + B)^3 = A^3 + B^3 + 3AB(A+B)$ .

Substituting the values and solving the equation for $A^3 + B^3$ , we get the answer as $\boxed {52}$ .

We have the equation $x^{2}-4x+1=0$ with roots A and B.

We know, $A+B=$ (-(coefficient of x))/(coefficient of $x^{2})$ $= \frac{-(-4)}{1} = 4$

Also, $AB=$ (constant term of the polynomial)/(coefficient of $x^{2})$ $= \frac{1}{1} = 1$

Now, $A^{3}+B^{3} = (A+B)(A^{2}-AB+B^{2}) = (A+B)((A+B)^{2}-3AB) = 4(4^{2}-3\times 1) = 4(16-3) = 4\times 13 =\boxed{52}$

A and B are roots of eqn. $x^{2}$ -4x+1=0

So, A+B= ${4/1}$

AB=1

$A^{3}$ + $B^{3}$ = $(A+B)^{3}$ -3 AB (A+B)

Putting value of A+B and AB we get $A^{3}$ + $B^{3}$ =52

Since it is a quadratic equation,

The sum of the roots i.e. A+B= -(coefficient of x)/(coefficient of x^2)

Therefore, A+B= -(-4)= 4

The product of the roots i.e. A*B = (constant)/(coefficient of x^2)

Therefore, A*B = 1

Since we need A^3 + B^3 that is equal to (A + B)(A^2 -A*B + B^2), we need to find the value of A^2 + B^2.

We know that, (A + B)^2 = A^2 +2 A*B + B^2

Therefore, A^2 + B^2 = (A + B)^2 - 2 A*B

Putting the values, we get

A^2 + B^2 = (4)^2 - 2*1

= 16 - 2

= 14

Now, A^3 + B^3 = (A + B)(A^2 -A*B + B^2)

= (A + B)(A^2 + B^2 -A*B)

= 4(14 - 1)

= 4*13

= 52

$\color{#20A900}{A^3+B^3=(A+B)^3-3AB(A+B)}$ From Vieta's formulas we know that $\color{#D61F06}{A+B=4,AB=1}$ .So $\color{#3D99F6}{A^3+B^3=4^3-3(1)(4)=64-12=\boxed{52}}$

according to the quadratic equation x=(-b±√(b^2-4ac))/2a. x = 2+3^1/2 (A) and 2-3^1/2 (B) . let them be equal to A and B. then solving by the identities (a+b)^3 = a^3 + b^3 + 3ab(a+b) and (a-b)^3 = a^3 - b^3 - 3ab(a-b) ,, we gt A^3 + B^3 = 52

If A and B are the roots of the given equation then A + B = 4 and A B =1 Then A^{3} + B^{3} = (A + B )^{3} - 3AB(A + B) = 4^{3} - 3 1*4 = 64 -12 = 52

IF P & Q ARE ROOTS OF A QUADRATIC EXPRESSION ax^2+bx+c=0 THEN P+Q=-b/a AND P*Q=c/a A+B= -b/a = - (- 4)/1 = 4 and AB = c/a = 1/1 = 1 hence, A^3 + b^3= (A+B)^3 - 3 AB (A+B) => 4^3 - 3 (1) (4) => 64 - 12 => 52

let A+B = -b/a and AB = c/a. and A^3 + b^3 = (A+B)^3 -3AB(A+B) so (4)^3 -3(1)(4) = 52

alpha + beta are the roots of quadratic equation ( class 9 th syllabus )

alpha + beta = -b/a = 4 alpha * beta = c/a = 1 (alpha + beta )^3 = alpha^3 + beta^3 + 3 alphta * beta ( alpha + beta ) [[a+b)^3 = a^3 + b^3 + 3ab(a+b)]]

=4^3 + 3 (1) (-4) = 64 -12

We know that if A and B are the roots of a quadratic ax^{2} + bx + c then, A+B= \frac{-b}{a} & AB= \frac{c}{a} . So, A+B=4 and AB=1. And we know that (A+B)^{3} = A^{3} + B^{3} + 3AB(A+B) or A^{3} + B^{3} = (A+B)^{3} - 3AB(A+B). Substituting the values of (A+B) and (AB) in the equation, A^{3} + B^{3} = 4^{3} - 3(1)(4) = 64-12 = \boxed{52}

A=2+sqt3 B=2-sqt3 A^3+B^3=52

A and B are the roots of the quadratic equation x2 - 4x + 1 = 0 <=> A + B = 4, A x B = 1. A^3 + B^3 = (A + B)(A^2 - A x B + B^2) = (A + B)[(A + B)^2 - 3x A x B] = 4 x ( 4^2 - 3 x 1) = 52

A+B= 4 AB=1 since, (A+B)^3= A^3 +B^3 + 3AB(A+B) this implies, A^3 + B^3 = (A+B)^3- 3AB(A+B) or A^3 + B^3 = 4^3 -3(4) or A^3 + B^3 = 52

Sum of the roots A+B =-b/a = 4 and Product of the roots AB =c/a = 1

Then we use A^3 +B^3 = (A+B)^3 -2AB(A+B) = (4)^3 -3 1 4 = 52

from given equation,we get A+B=4 and AB=1 ( as given roots are A and B ) ,So we get A^3+B^3= 4^3-3 1 4=52

If $A$ and $B$ are the roots of the equation $x^2-4x+1=0$ then the equation can be written as $(x-A)(x-B)=0 \iff x^2-(A+B)x+AB=0$ . Therefore $A+B=4$ .

That $A^2-4A+1=0$ and $B^2-4B+1=0$ is true since $A$ and $B$ are the roots of the equation. By addition, $(A^2+B^2)-4(A+B)+2=0$ , so $A^2+B^2=4(A+B)-2=14$ .

Also $A^3-4A^2+A=0$ and $B^3-4B^2+B=0$ , by addition again, $(A^3+B^3)-4(A^2+B^2)+(A+B)=0$ , so $A^3+B^3=4(A^2+B^2)-(A+B)=52$ .