Math, asked by SmartAkk, 1 year ago

Solve the following quadratic equations by factorisation method

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Answered by ritujain83

Answer:

Step-by-step explanation:

x² + 3 x - ( a² + a - 2 ) = 0

==> x² + 3 x - ( a² + 2 a - a - 2 ) = 0

==> x² + 3 x - ( a ( a + 2 ) - 1 ( a + 2 ) ) = 0

==> x² + 3 x - ( a + 2 )( a - 1 ) = 0

==> x² + ( ax + 2 x - ax + x ) - ( a + 2 )( a - 1 ) = 0

==> x² + x( a + 2 ) - x ( a - 1 ) - ( a + 2 )( a - 1 ) = 0

==> x ( x + a + 2 ) - ( a - 1 ) ( x + a + 2 )

==> ( x + a + 2 )( x - a + 1 ) =0

2.....

x²-4ax+4a²-b²=0

==>x²-2*2a*x+(2a)²-(b)²=0

==>(x-2a)²-(b)²=0

==>(x-2a-b)(x-2a+b)=0

Hope this helps u..

Please mark the brainliest...

SmartAkk: where is the root of the equation

rcagindiap9s19c: Friend, I have answered your question and yes the root of the equation is the answer of the equation.

ritujain83: That's a simple task..

ritujain83: That can be done by equating the factors to zero.. that must be known...

ritujain83: Copying the answer from somewhere and pasting it here is a simple thing.. anyone can do it.. BUT WHEN TYPING IT WHOLE, THEN ANSWERING.. THEN ALSO NOT GETTING THE WHAT U DESERVE IS LIKE BEING CHEATED....

SmartAkk: behave properly

ritujain83: What???

Answered by rcagindiap9s19c

Step-by-step explanation:

$(i. )x^2+3x-(a^2+a-2) = 0\\= x(x+3)-(a^2 +2a -a -2) = 0\\=x(x+3)(a^{2} + 2a -a -2)= 0\\ =x(x+3)a(a+2)-1(a+2) = 0\\=x(x+3)(a-1)(a+2)=0\\ =x = 0\\or (x+3)=0\\x = -3\\or (a-1)(a+2)=0\\a = 1\\ or a=-2\\Ans= x = 0 \\or -3\\and a = 1 \\or -2\\\\\\$

ii.) x^2 - 4ax - 4a^2 - b^2

= $x^2 - 2 * x * 2a + (2a)^2 - b^2= 0\\= (x-2a)^2 - b^2 = 0\\= (x-2a+b)(x-2a-b) = 0\\= x = 2a -b\\or\\x=2a+b$

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