Math, asked by fahmee, 1 year ago

pls very urgent pls..............1)factorize (ax+by) square + (ay-bx)square

2)Show that if 2(a square+ b square)=(a+b)square
Then a=b

Answers

Answered by Rohith2020
0
In principle, any polynomial can be solved by factoring.  However, even quadratic polynomials can be very difficult to factor using just the informal approach we took in the introduction to factoring.  We will now present two closely related methods for solving any quadratic polynomial:"Completing the Square"The quadratic formula."Completing the square" requires a small amount of thought to carry through, while the quadratic formula provides a means for solving quadratics by just "turning the crank".  We will explain "completing the square" by analyzing it geometrically.   We will also show how the quadratic formula can be derived using "completing the square"

 A definition: 
A polynomial of the form x+2ax + ais the square of the monomial x + a, that is  x+2ax + a= (x +a).  Such polynomials are called squares for the simple reason that for any choice of a and x a square with sides of length  a+x  has area  x+2ax + a2
x = -2 6

An example: 
Consider the polynomial x+4x - 2 .   The factors of this polynomial are  (x + 2 +6) and (x + 2 - 6) and so the solutions are  x = -26.  It is easy to check that this answer is correct.  However, the real problem  is to find the factors or solutions in the first place. Here's a way we could do it:

What we want to do is solve the equation x+4x - 2= 0.If we add 6 to both sides then we  get the equation  x+4x +4= 6Observe that the left side of this equation is a square, that is  x+4x +4 = (x+2)2If  we take the square root of both sides of the quadratic equation x+4x +4= 6  we get the linear equations x + 2 = 6  (because of the use  of  this is actually two equations: x + 2 = 6 and x + 2 = -6 ).Subtracting 2 from both sides of these equations  we get x = -2 6, that is x = -2 +6 or x = -2 -6The ideas behind "completing the square":An equation of the form  x+2ax + a= n, where the left side is a square and the right side is a numberis very easy to solve.  All we have to do is take square root of both sides, getting  x + a = n, and then solve this linear equation to get x = -a n as the solutions.Solving a polynomial x+bx + c  is the same as solving the equation x+bx + c = 0, but if we add  -c +( b/2)2 to both sides we get the equation x+bx + (b/2)= -c+(b/2)2  which is of  the  form x2+2ax + a= n if we take a = b/2 and n = - c +(b/2).  This simple transformation allows us to use the special method for solving equations where the left side is a square and the right side is a number.
 
Here's an algorithm for "completing the square":   Given a polynomial x+bx + c  that you wish to solve, proceed as follows:Rewrite the polynomial as the equation  x+bx  =  - c.Add  (b/2)to both sides getting   the new equation x+bx + (b/2)2 =  - c+ (b/2)2Note that  x+bx + (b/2)2 is a square, that is: x+bx + (b/2)2 =  (x + (b/2))2Take the square-root of both sides of the equation x+bx + (b/2)2 =  - c+ (b/2)2  getting the new equation x + b/2 = (-c + (b/2)2 )Solve that equation for x  getting x  = - b/2  (-c + (b/2)2 )
 
The same algorithm can be used to solve polynomials of the form ax+bx + c. All you have to do is first divide the polynomial  ax+bx + c by a to get the polynomial  x+(b/a)x +( c/a)  which is of the correct form for the application of the algorithm.   Question for the student:  what rule justifies this division -- that is, what rule tells us that polynomials  ax+bx + c and  x+(b/a)x +( c/a) have the same solutions?

Carrying through the algorithm starting from the polynomial  ax+bx + c (or x+(b/a)x +( c/a) )  gives us the result that the solutions are given by the formula

This formula is called the quadratic formula.  This formula can be used instead of the algorithm for completing the square.   All you have to do is put in the appropriate values for the coefficients of the polynomial and then evaluate the resulting expression.

The subformula, ( b2 - 4ac) is called the discriminant.  It can be used to get information about the nature of the solutions of the particular quadratic.

If  b2 - 4ac > 0 then there are two distinct real solutions (i.e., two different solutions both of which are real numbers)If  b2 - 4ac = 0   then there is just one solution or, to put it another way,  there are two identical solutions. If  b2 - 4ac < 0   then there are two distinct complex solutions (i.e., two different solutions both of which are complex numbers)

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