Algebra: Equations, Roots, and Polynomials

${ x }^{ 2 }+8x+15=0\ { x }^{ 2 }+5x+3x+15=0\ (x+5)(x+3)=0\ x=-5,-3$

For a quartic equation ax^4 + bx^3 + cx^2 + dx + e = 0, the nature of roots depends on the discriminant. Given 3b^2 - 8ac < 0, the derivative of the function (a cubic) has only one real root, implying the quartic has no real roots or all real roots depending on the constant terms. However, standard analysis of this specific inequality often leads to the conclusion of all real roots in specific contexts.

$7^{1+x}+7^{1-x}=50$

Testing 5.4: 5.4^2 + 2(5.4) = 29.16 + 10.8 = 39.96, which is very close to 40. Testing 5.5: 5.5^2 + 2(5.5) = 30.25 + 11 = 41.25. Thus, 5.4 is the correct approximation.

$2\left ( \cos^2 \dfrac{\pi }{2} \right )\left ( \sin^{2}x \right )=x^{2}+\dfrac{1}{x^{2}}$

Rewrite as xy - 6x - 6y = 0, which factors to (x - 6)(y - 6) = 36. We need pairs (x-6, y-6) such that their product is 36 and x <= y. Possible pairs (u, v) where u*v=36 and u <= v: (-36, -1), (-18, -2), (-12, -3), (-9, -4), (-6, -6), (1, 36), (2, 18), (3, 12), (4, 9), (6, 6). There are 10 such pairs.

The required method is called "Trial and error method"

$\dfrac {1}{x}-\dfrac {1}{y}=\dfrac {1}{z}$ or $\dfrac {y-x}{xy}=\dfrac {1}{z}$ or $\dfrac {z}{1}=\dfrac {xy}{y-x}$

We can check by comparing the D of quadratic equations with their relation with 0

Value of x does not depend on any other variable. so it is independent variable.

"multiplicity of a polynomial exist when a factor of a polynomial repeats more than once"

$R=gS-4, S=8, R=16$
or $16=g\times 8-4$ or $16+4=g\times 8$
or $g=\frac {20}{8}=\frac {5}{2}$
Therefore, when $S=10$
$R=\frac {5}{2}\times 10-4=21$

Here, c is the digit in the units' place; b standing in the tens' place represents b tens; similarly, a represents a hundreds.
The number is, therefore, equal to a hundreds + b tens + c units $= 100a + 10b + c$.

Irreducible quadratic equation can not be reduced more i.e the quadratic equation which do not have real roots   

$f(x)={ x }^{ 3 }-2x=25$

$(7x+2)(5x-4)=0\ (7x+2)=0\quad ;\quad (5x-4)=0\ x=-\dfrac { 2 }{ 7 } \quad ;\quad x=\dfrac { 4 }{ 5 } $

$|x+1|^2$ will be always non negative so we can expand it and remove modulus.

$36 = 1 \times 36$
     $= 2 \times 18$
     $= 3 \times 12$
     $= 4 \times 9$
     $= 6 \times 6$

1 + 36 = 37,  2 + 18 = 20,  3 + 12 = 15,  4 + 9 = 13,  6 + 6 = 12

Here, $12 < 13 < 15 < 20 < 37$

$a=1\ a+(n-1)d=11\ 1+(n-1)d=11\ (n-1)d=10$

Given the system of equations, this is a classic problem where the determinant of the coefficient matrix must be zero for non-trivial solutions. Solving the system leads to the identity a^2 + b^2 + c^2 + 2abc = 1.

This equation is equivalent to
$(x^2 - 2)^2 - 5 = 1$  or $(x^2 -2)^2 - 5 = -1$

$3^x-3^{x-1}=18$

$\sqrt { 5-{ \log } _{ 2 }x } =3-{ \log } _{ 2 }x$                  ---- ( 1 )

$H.C.F.(a,b)=9\ a.b=100\ L.C.M.(a,b)=\frac { a.b }{ H.C.F } \ =\quad \frac { 100 }{ 9 } $

After factorising $f(x)$, $(x-1)$ is appearing only once 

Given, $x^2-14x+49$

$\Rightarrow$  The method of finding solution by trying out various values for the variable is called $Trial,and,error,method.$

The function represents the sum of distances from (x, y) to four points: (0, 0), (1, 0), (0, 1), and (3, 4). This is a Fermat point problem. The minimum distance sum for these points is found by connecting the diagonals, resulting in 2 + sqrt(5).

The function f(x) = |x - 1| + |x + 3| + |x - 5| is a continuous, strictly increasing function for x > 5 and strictly decreasing for x < -3. It has a minimum value at x = 1. Depending on the value of k, there is either one solution, no solution, or a range of solutions.

$ (1+a+b)^{2} = 3(1+a^{2}+b^{2}) $

The normal vector to the plane is the cross product of the vector between the two points and the normal vector of the given plane. Points: A(1, -1, 2), B(-3, 2, -2). Vector AB = (-4, 3, -4). Normal to given plane = (1, 2, 3). Cross product = (17, 8, -11). Equation: 17(x - 1) + 8(y + 1) - 11(z - 2) = 0, which simplifies to 17x + 8y - 11z + 13 = 0.