3x 2 X 2 1

Dr.Dipin Singh 8 years Ago

(ten-i)(x-ii)(3x-2)(3x+1) = 21
(ten-i)*(3x-2)(3x+ane)*(10-two) = 21
(3x^ii-5x+two)(3x^2-5x-2)=21
put 3x^2-5x=y
(y+two)*(y-2)=21
y^ii=25
y= +-5
put y= 3x^2-5x=5
3x^2-5x-v=0
and then
x= (5+Sqrt 85)/half dozen, (5-Sqrt 85)/half-dozen ... selection 7

Similar?Yes (17) | No (3)

Ann Theressa eight years Ago

(x-1)(x-2)(3x-ii)(3x+1) = 21
(x-ane)(3x-2)(x-2)(3x+i) = 21
(3x^2-5x+2)(3x^2-5x-two) = 21

put 3x^2 - 5x = y

(y+2)(y-2) = 21
y^two - iv = 21
y^2 = 25
y = (+-)5

3x^two - 5x = 5
3x^2 - 5x -5 = 0

ten = {5(+-)sqrt(25+threescore)}/6
=(5+Sqrt 85)/half-dozen, (5-Sqrt 85)/6

Like?Yeah (12) | No (5)

Dada Khalandar eight years AGO

(x-1)(x-2)(3x-2)(3x+1) = 21

9x^four - 30x^3 + 25x^ii - iv = 21

9x^4 - 30x^3 + 25x^2 = 25

(3x^2 - 5x)^2 = 25

3x^2 - 5x - 5 = 0 or 3x^2 - 5x + 5 = 0

x = (v ± √85) / half-dozen or x = (5 ± i√35) / vi

Like?Yes (vii) | No (2)

MD SOHAIL AKHTER 8 years AGO

(ten-1)(x-2)(3x-2)(3x+ane) = 21
(3x^2-5x+2)(3x^ii-5x-2)= 21
(3x^2-5x)^2= 25
(3x^ii-5x)^ii-5^ii=0
(3x^2-5x-5)(3x^2-5x+5)=0

on solving, we get
ten=(five+Sqrt 85)/6, (5-Sqrt 85)/half dozen

o choice (7) is correct

Like?Yes (v) | No (iii)

Udhaya kumar viii years Ago

(x-1)(3x-two)(x-two)(3x+1)=21
by solving 3x^2-5x-five=0
ten=5+sqrt85/6

Like?Yes (4) | No (3)

subasini matheswaran viii years AGO

9x^4 - 30x^3 + 25x^2 - 4 = 21

9x^four - 30x^3 + 25x^2 = 25

(3x^ii - 5x)^2 = 25

3x^2 - 5x - 5 = 0 or 3x^2 - 5x + 5 = 0

x = (5 - √85) / 6 , 10 = (five- srqt 85) / 6

Similar?Yes (2) | No (one)

Bhagyashree Bhatt 8 years Ago

offset expand:
(x - 1)(x - 2)(3x - 2)(3x + 1) = 21
(10² - 3x + ii)(9x² - 3x - 2) = 21
(9x^4 - 3x³ - 2x²) + (-27x³ + 9x² + 6x) + (18x² - 6x - 4) = 21
9x^4 - 30x³ + 25x² + 0x - iv = 21
take out common factor x² and movement constant to RHS:
x²(9x² - 30x + 25) = 25
the chip in parentheses is a square:
x²(3x - 5)² = 25
take square root of both sides:
x(3x - 5) = 5
rearrange:
3x² - 5x - 5 = 0
sub these numbers into the equation to solve for roots of quadratic:
x = (-b ± sqrt(b² - 4ac))/(2a)
= (5 ± sqrt(25 - (iv*3*-5))/(2*3)
= (5 ± sqrt(25 - (-60))/6
= (v ± sqrt(85))/half-dozen

Like?Yep (2) | No (two)

sanghavi 8 years AGO

9x^4-30x^3+25x^ii=25
x^2(3x-v)^2=25
ten(3x-five)=5
3x^2-5x-5=0
by solving this equation
we become option 7

Similar?Yes (1) | No (two)

AMIT SINGH eight years Agone

(10-1)(3x-two)(x-2)(3x+ane)=21
(3x^2-5x+2)(3x^2-5x-2)=seven*iii
on equating
3x^2-5x+2=7 and 3x^2-5x-2=iii
we have,
3x^ii-5x-5=0
hence, on solving
ten=(5+sqrt85)/6,(five-sqrt85)/6.

Like?Yes (2) | No (3)

sruthi eight years AGO

7 th one is the reply

Like?Yep (one) | No (two)

Sushma 8 years AGO

Regroup the factors.
put 3x^2-5x=y
solve new quadratic eqn to get y= +-5.
solve 3x^2-5x=y=v to become the upshot.

Like?Yes (two) | No (3)

srilakshmi 8 years Agone

(x-1)(10-2)(3x-two)(3x+one) = 21
(ten-1)(3x-2)(x-2)(3x+1) = 21
(3x^2-5x+2)(3x^2-5x-2) = 21
let us assume 3x^ii-5x = x
past solving the above we will become
3x^2-5x-5= 0
roots are
-b+(or)-Sqrt(b^2-4ac)/2a
(five+Sqrt 85)/half dozen,(5-Sqrt 85)/half-dozen

Like?Yep (ane) | No (3)

T.Aravinth kumar 8 years AGO

Sub options in equation

Similar?Yes (1) | No (three)

deepak kumar 8 years Ago

normal calculation by quadratic equation.

Similar?Yes (1) | No (3)

rajani 8 years AGO

(x-1)(3x-2)(x-2)(3x+ane)=21
->(3x^2-5x+2)(3x^two-5x-two)=21
allow 3x^2-5x be 'a'
(a+2)(a-two)=21
a^2-4=21
->a^2=25
->a=+5 or -5
Past taking the root every bit -5,
eq is 3x^2-5x=-5
on solving this nosotros get (v+sqrt85)/half dozen and (5-sqrt 85)/6

Like?Yes (1) | No (four)

Gitanjali 1000 8 years Ago

(10 - ane)(x - ii)(3x - 2)(3x + 1) = 21
(ten^two - 3x + 2)(9x^ii - 3x - 2) = 21
9x^iv - 30x^iii + 25x^two - 4 = 21
9x^four - 30x^iii + 25x^2 = 25
10^2(9x^2 - 30x + 25) = 25
10^2(3x - 5)^ii = 25

accept square root of both sides:
x(3x - v) = v
3x^two - 5x - 5 = 0

solve quadratic equation using

x = (-b (+/-) sqrt(b^2 - 4ac))/(2a)
= (five (+/-) sqrt(25 - (-60))/half dozen
= (5 (+/-) sqrt(85))/vi
therefore
x=(5+Sqrt 85)/half-dozen, x=(five-Sqrt 85)/6
so option (seven) is the answer

Like?Yep (1) | No (four)

Anupam Banerjee 8 years AGO

(10-1)(x-ii)(3x-2)(3x+1)=21
or, (ten-one)(3x-ii)(x-2)(3x+1)=21
or, (3x^ii -5x +2)(3x^2 -5x -2)=21

Let, 3x^two-5x=z;
Then, (z+2)(z-2)=21
or, z^2 - 4=21
or, z=five;

So, 3x^ii - 5x = 5
or, 3x^2 - 5x - 5=0
or, x=(5+Sqrt(25+sixty))/6,
ten=(5-Sqrt(25+lx))/6;

Like?Aye (1) | No (4)

3x 2 X 2 1,

Source: https://m4maths.com/previous-puzzles-1304-Solve-x-1-x-2-3x-2-3x-1-21.html

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