1.

find the zeroes of polynomial by Quadratic formula ​

Answer» ONG>Step-by-step explanation:

2(2a^2+5ab+2b^2)=4a^2+10ab+<klux>4B</klux>^2\\=(2a)^2+2a(5b)+4b^2\\

So we NEED two numbers which add to 5b and multiply to 4b^2

Which are : 4b and b

therefore:

2(2a^2+5ab+2b^2)=(2a)^2+2a(5b)+4b^2=(2a+4b)(2a+b)\\=>(2a^2+5ab+2b^2)=\frac{(2a+4b)(2a+b)}{2} =\frac{2(a+2b)(2a+b)}{2}=(a+2b)(2a+b)\\=-(a+2b)*-(2a+b)=(-a-2b)(-2a-b) \\therefore: 9x^2-9(a+b)+(2a^2+5ab+2b^2)\\\\=(3x)^2+(3x)(-3)(a+b)+ (-a-2b)(-2a-b)\\=(3x)^2+(3x)(-3a-3b)+ (-a-2b)(-2a-b)

since (-a-2b) and (-2a-b) add up to (-3a-3b), we can factorize the polynomial like this:

(3x-a-2b)(3x-2a-b)=0=>\\3x = a+2b=>x=\frac{a+2b}{3} \\3x=2a+b=>x=\frac{2a+b}{3}


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