IIT JEE , AIEEE Forum - Mathematics , Algebra

sneha_91

(x^2-4x)^2 -(x-4)^2-16=0

please i want this answer immediately..............................

magiclko

JimboJones

Equation: (x² - 4x)² - (x - 2)² - 16 = 0

ð (x² - 4x + 8 - 8)² - (x - 2)² - 16 = 0

ð ( ( x - 2)² - 8)² - (x - 2)² - 16 = 0 - (I)

Substitute: y = (x - 2) ²

Thus (I) becomes

ð ( y - 8)² - y - 16 = 0

ð y² - 17y + 48 = 0

ð y = ( 17 ± sqrt( 17* 17 - 4 * 48) ) / 2

ð y = ( 17 ± sqrt(97) ) / 2

ð (x - 2) ² = ( 17 ± sqrt(97) ) / 2

ð x - 2 = ± ( 17 ± sqrt(97) ) / 2

ð x = 2 ± ( ( 17 ± sqrt(97) ) / 2 )

Thus the equation has 4 solutions. They are.

1. 2 + ( ( 17 + sqrt (97) ) / 2 )

2. 2 + ( ( 17 - sqrt (97) ) / 2 )

3. 2 - ( ( 17 + sqrt (97) ) / 2 )

4. 2 - ( ( 17 - sqrt (97) ) / 2 )

amar.gupta

Dear,

it is not possible to always write each polynomial into the factor form.

however if you want to solve this .on resolving the eq. you will get:

f(x)= x^4 - 8x^3 + 15x^2 + 8x - 32 = 0

now let f(x) =(x^2 +ax+ b)(x^2 +cx +d)

resolve it and compare the coeff. of corresponding power of x. you will left with four eq. solve it and you get the two quadratics . those you can solve easily.

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