M4maths Previous Puzzles

15 May 2014 Quadratic Equation

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

OPtion

1) (4+Sqrt 85)/6, (4-Sqrt 85)/6
2) (3+Sqrt 75)/6, (3-Sqrt 75)/6
3) (5+Sqrt 75)/6, (5-Sqrt 75)/6
4) (7+Sqrt 7)/6, (7-Sqrt 7)/6
5) (5+Sqrt 5)/6, (5-Sqrt 5)/6
6) (3+Sqrt 26)/6, (3-Sqrt 26)/6
7) (5+Sqrt 85)/6, (5-Sqrt 85)/6
8) (4+Sqrt 5)/6, (4-Sqrt 5)/6
9) (5+Sqrt 10)/6, (5-Sqrt 10)/6
10)None of these

Solution

Solve
Take 3 as common factor from last two factor to get
9(x-1)(x-2)(x- 2/3)(x+ 1/3) = 21
The sum of constant terms of first factor and third factor is same as the sum of the constant terms of second and fourth factor.
(3x^2-5x+2)(3x^2-5x-2) = 21
let 3x^2-5x = t
(t+2)(t-2) = 21
Solve

Answer is (5+Sqrt 85)/6, 5-Sqrt 85)/6

Option 7)

Correct Option: 7 Best Solution (17)

Dr.Dipin Singh   10 years AGO

(x-1)(x-2)(3x-2)(3x+1) = 21
(x-1)*(3x-2)(3x+1)*(x-2) = 21
(3x^2-5x+2)(3x^2-5x-2)=21
put 3x^2-5x=y
(y+2)*(y-2)=21
y^2=25
y= +-5
put y= 3x^2-5x=5
3x^2-5x-5=0
then
x= (5+Sqrt 85)/6, (5-Sqrt 85)/6 ... option 7

Like? Yes (17) | No (3)  

Ann Theressa   10 years AGO

(x-1)(x-2)(3x-2)(3x+1) = 21
(x-1)(3x-2)(x-2)(3x+1) = 21
(3x^2-5x+2)(3x^2-5x-2) = 21

put 3x^2 - 5x = y

(y+2)(y-2) = 21
y^2 - 4 = 21
y^2 = 25
y = (+-)5

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

x = {5(+-)sqrt(25+60)}/6
=(5+Sqrt 85)/6, (5-Sqrt 85)/6

Like? Yes (12) | No (5)  

Dada Khalandar   10 years AGO

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

9x^4 - 30x^3 + 25x^2 - 4 = 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 = (5 ± √85) / 6 or x = (5 ± i√35) / 6

Like? Yes (7) | No (2)  

MD SOHAIL AKHTER   10 years AGO

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

on solving, we get
x=(5+Sqrt 85)/6, (5-Sqrt 85)/6

o option (7) is correct

Like? Yes (5) | No (3)  

Udhaya kumar   10 years AGO

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

Like? Yes (4) | No (3)  

subasini matheswaran   10 years AGO

9x^4 - 30x^3 + 25x^2 - 4 = 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 = (5 - √85) / 6 , x = (5- srqt 85) / 6

Like? Yes (2) | No (1)  

Bhagyashree Bhatt   10 years AGO

first expand:
(x - 1)(x - 2)(3x - 2)(3x + 1) = 21
(x² - 3x + 2)(9x² - 3x - 2) = 21
(9x^4 - 3x³ - 2x²) + (-27x³ + 9x² + 6x) + (18x² - 6x - 4) = 21
9x^4 - 30x³ + 25x² + 0x - 4 = 21
take out common factor x² and move constant to RHS:
x²(9x² - 30x + 25) = 25
the bit 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 - (4*3*-5))/(2*3)
= (5 ± sqrt(25 - (-60))/6
= (5 ± sqrt(85))/6

Like? Yes (2) | No (2)  

sanghavi    10 years AGO

9x^4-30x^3+25x^2=25
x^2(3x-5)^2=25
x(3x-5)=5
3x^2-5x-5=0
by solving this equation
we get option 7

Like? Yes (1) | No (2)  

AMIT SINGH   10 years AGO

(x-1)(3x-2)(x-2)(3x+1)=21
(3x^2-5x+2)(3x^2-5x-2)=7*3
on equating
3x^2-5x+2=7 and 3x^2-5x-2=3
we have,
3x^2-5x-5=0
hence, on solving
x=(5+sqrt85)/6,(5-sqrt85)/6.

Like? Yes (2) | No (3)  

sruthi   10 years AGO

7 th one is the answer

Like? Yes (1) | No (2)  

Sushma   10 years AGO

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

Like? Yes (2) | No (3)  

srilakshmi   10 years AGO

(x-1)(x-2)(3x-2)(3x+1) = 21
(x-1)(3x-2)(x-2)(3x+1) = 21
(3x^2-5x+2)(3x^2-5x-2) = 21
let us assume 3x^2-5x = x
by solving the above we will get
3x^2-5x-5= 0
roots are
-b+(or)-Sqrt(b^2-4ac)/2a
(5+Sqrt 85)/6,(5-Sqrt 85)/6

Like? Yes (1) | No (3)  

T.Aravinth kumar   10 years AGO

Sub options in equation

Like? Yes (1) | No (3)  

deepak kumar   10 years AGO

normal calculation by quadratic equation.

Like? Yes (1) | No (3)  

rajani   10 years AGO

(x-1)(3x-2)(x-2)(3x+1)=21
->(3x^2-5x+2)(3x^2-5x-2)=21
let 3x^2-5x be 'a'
(a+2)(a-2)=21
a^2-4=21
->a^2=25
->a=+5 or -5
By taking the root as -5,
eq is 3x^2-5x=-5
on solving this we get (5+sqrt85)/6 and (5-sqrt 85)/6

Like? Yes (1) | No (4)  

Gitanjali K   10 years AGO

(x - 1)(x - 2)(3x - 2)(3x + 1) = 21
(x^2 - 3x + 2)(9x^2 - 3x - 2) = 21
9x^4 - 30x^3 + 25x^2 - 4 = 21
9x^4 - 30x^3 + 25x^2 = 25
x^2(9x^2 - 30x + 25) = 25
x^2(3x - 5)^2 = 25

take square root of both sides:
x(3x - 5) = 5
3x^2 - 5x - 5 = 0

solve quadratic equation using

x = (-b (+/-) sqrt(b^2 - 4ac))/(2a)
= (5 (+/-) sqrt(25 - (-60))/6
= (5 (+/-) sqrt(85))/6
therefore
x=(5+Sqrt 85)/6, x=(5-Sqrt 85)/6
so option (7) is the answer

Like? Yes (1) | No (4)  

Anupam Banerjee   10 years AGO

(x-1)(x-2)(3x-2)(3x+1)=21
or, (x-1)(3x-2)(x-2)(3x+1)=21
or, (3x^2 -5x +2)(3x^2 -5x -2)=21

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

So, 3x^2 - 5x = 5
or, 3x^2 - 5x - 5=0
or, x=(5+Sqrt(25+60))/6,
x=(5-Sqrt(25+60))/6;

Like? Yes (1) | No (4)