Pls factorise this. Its urgent.
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Given, 4x^2 - 4a^2x + (a^4-b^4) = 0
Formula is -b+ root of b^2-4ac/2a.
After substituting values in above equation, we get
= 4a^2 + root 16a^4 - 4*4*(a^4-b^4)/2 * 4
= 4a^2 + root of 16b^4/8
= 4a^2 +4b^2/8
= 4(a^2+b^2)/2 * 4
= a^2 + b^2/2
So the values are a^2 + b^2/2 and a^2-b^2/2.
Hope this helps!
Formula is -b+ root of b^2-4ac/2a.
After substituting values in above equation, we get
= 4a^2 + root 16a^4 - 4*4*(a^4-b^4)/2 * 4
= 4a^2 + root of 16b^4/8
= 4a^2 +4b^2/8
= 4(a^2+b^2)/2 * 4
= a^2 + b^2/2
So the values are a^2 + b^2/2 and a^2-b^2/2.
Hope this helps!
Kushagrasingh001:
Bro thanks for help, but the question was to find by factorisation method.
Alright then, Wait for 2 minutes.
Given, 4x²-4a²x+(a⁴-b⁴)=0
= 4x²-2(a²+b²)+(a²-b²)x+(a²+b²)(a²-b²)=0
= 2x(2x-(a²+b²))-(a²-b²)(2x-(a²+b²))=0
= (2x-(a²+b²))(2x-(a²-b²))=0
= a^2+b^2/2, a^2-b^2/2
= 4x²-2(a²+b²)+(a²-b²)x+(a²+b²)(a²-b²)=0
= 2x(2x-(a²+b²))-(a²-b²)(2x-(a²+b²))=0
= (2x-(a²+b²))(2x-(a²-b²))=0
= a^2+b^2/2, a^2-b^2/2
Thnk u vry much.
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