Question

4 x square minus y square + 6 Y - 9 factorise​

Answer 1

$\huge\mathbb{SOLUTION:-}$

$\sf\underline{\underline{\purple{Given:-}}}$

$\sf :\implies 4x{}^{2}-y{}^{2}+6y-9$

We have to factorize this

Now formula used

$\boxed{\sf{(a-b){}^{2}=a{}^{2}-2ab+b{}^{2}}}$

$\boxed{\sf{a{}^{2}-b{}^{2}=(a+b)(a-b)}}$

$\sf\underline{\underline{\red{A.T.Q:-}}}$

$\sf:\implies 4x{}^{2}-y{}^{2}+6y-9\\ \\ \sf:\implies 4x{}^{2} -(y{}^{2}-6y+9)\\ \\ \sf:\leadsto (y-3){}^{2}=y{}^{2}-6y+9\\ \\ \sf:\implies 4x{}^{2}-(y-3){}^{2}\\ \\ \sf:\leadsto this\:is\:in\:form\:of\:a{}^{2}-b{}^{2}\\ \\ \sf:\implies (2x){}^{2}-(y-3){}^{2}\\ \\ \sf:\implies [2x+(y-3)][2x-(y-3)]\\ \\ \sf:\implies (2x+y-3)(2x-y+3)$

$\boxed{\sf{(2x+y-3)(2x+y-3)}}$

Answer 2

Answer:

[2x-y+3]×[2x+y-3] = 0

Step-by-step explanation:

4x²-y²+6y-9 = 0

0 = 4x²- (y²-6y+9)

0 = 4x²- [(y)² - 2(y)(3) + (3)²]

0 = 4x²- [y-3]² ...... { (a²-2ab+b²) = (a-b)² }

0 = (2x)² - (y-3)²

0 = [{2x - (y-3)]×[{2x + (y-3)] ....... { a²-b² = (a+b)×(a-b) }

0 = [2x-y+3]×[2x+y-3]

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