# Find the value of a for which one root of the quadratic equation (a2-5a+3)x2+(3a-1)x+2=0 is twice as large as the other.How can we eliminate alpha here? Please solve the problem?

Let one root be T then the other root will be 2T.

Comparing equation (a² - 5a + 3)x² + (3a - 1)x + 2 = 0 with Ax² + Bx + C =0

We get A = (a² - 5a + 3), B = (3a - 1) ,  C = 2

Sum of roots =
3T =   (1)

And product of roots = 2T2 (2)
Comparing (1) and (2) , we have

Hence if we put a = 2/3
Then the roots will be -3 and -6 .

• 80

Let one root be T then the other root will be 2T.

Comparing equation (a² - 5a + 3)x² + (3a - 1)x + 2 = 0 with Ax² + Bx + C =0

We get A = (a² - 5a + 3), B = (3a - 1) C = 2

Sum of roots = - B/A Putting values we get

T + 2T = -(3a - 1)/(a² - 5a + 3) = 3T = - ( 3a - 1)/(a² - 5a + 3)

= T = - (3a - 1) /3(a² - 5a + 3).........(1)

Also product of roots = C/A putting values again

T x 2T = 2/(a² - 5a + 3) = 2T² = 2/(a² - 5a + 3)

= T² = 2/2(a² - 5a + 3) = T² = 1/(a² - 5a + 3)

= { -(3a - 1)/3(a² - 5a + 3) }² = 1/ (a² - 5a + 3) ( Putting value of T from equation (1) )

= {(3a - 1)²/9(a² - 5a +3)²} X (a² - 5a + 3) = 1 { Multiplying both sides by (a² - 5a + 3) }

= (3a - 1)²/9(a² - 5a + 3) = 1

= (3a -1)² = 9(a² - 5a + 3)

= (3a)² + (1)² - 2 (3a) (1) = 9a² - 45a + 27 { Applying (a - b)² = a² + b² - 2ab }

= 9a² + 1 - 6a = 9a² - 45a + 27

= 9a² - 45a - 9a² + 6a = 1 - 27 ( Rearranging the equation)

= 9a² - 9a² - 45a + 6a = - 26 (Again rearranging the equation)

= - 39a = - 26 = a = -26/-39

= a = 2/3

• 22
What are you looking for?