Math, asked by vedantgolait, 1 year ago

Solve the following by factorization method
4x² - 2(a² + b²)x + a²b² = 0

Attachments:

Answers

Answered by ShuchiRecites
4

Equation: 4x² - 2(a² + b²)x + a²b²

So you know how to factorize ryt? First of all multiply 1st and last term :-)

→ 4x² × a²b²

→ 4x²a²b²

Now focus middle term, - 2(a² + b²)x

Let's unbox it, xD

→ - 2(a²x + b²x)

→ - 2a²x - 2b²x

Multiply both terms, (-2a²x • - 2b²x) = 4x²a²b².

Hence now let's start factorizing,

→ 4x² - 2a²x - 2b²x + a²b²

→ 2x(2x - a²) - b²(2x - a²)

→ (2x - a²)(2x - b²) = 0

Answer: x = a²/2 and b²/2

Answered by rishu6845
1

Answer:

x=a²/2 and x=b²/2

Step-by-step explanation:

To find --->Solve following equation by

------------ factorization method

4x²-2(a²+b²)x+a²b²=0,

Solution--->

-------------

4x²-2(a²+b²)x+a²b²=0

Multiplying x by 2(a²+b²) in middle term

=> 4x²-2(a²x+b²x)+ a²b² =0

Changing the sign of middle term

=> 4x²-2a²x-2b²x+a²b²=0

Taking 2x common from first two terms and (-b²) from last two terms

=> 2x(2x-a²)-b²(2x-a²)=0

Now taking (2x-a²) common

=> (2x-a²)(2x-b²)=0

If 2x-a²=0

=> 2x=a²

=> x=a²/2

If 2x-b²=0

=> 2x=b²

=> x=b²/2

So x=a²/2 or x=b²/2

Additional information--->

-------------------------------------

1) ax²+bx+c=0

Solution of quadertic equation is

-b ±√(b²-4ac)

x=-----------------------

2a

2) ax²+bx+c=0

If roots of equation is α and β then

α+β=-b/a

αβ=c/a

3) If roots of equation are α and β then equation is

x²-(sum of roots )x+(product of

roots)=0

Similar questions