find the value of a if one root of the equation 8×2-6x+a=0 is the square of the other​

2 thoughts on “find the value of a if one root of the equation 8×2-6x+a=0 is the square of the other​”

1. Answer:

   Mark as brainliest

   Step-by-step explanation:

   ⇒ The given quadratic equation is 8x

   2

   −6x+k=0, comparing it with ax

   2

   +bx+c.

   ⇒ Then, a=8,b=−6,c=k

   ⇒ It is given that one root of this equation is square of the other root. So, if we assume one root to be p, the other root can be p

   2

   . So, we assume the roots to be p and p

   2

   .

   ⇒ Sum of the roots =

   a

   −b

   ∴ p+p

   2

   =

   8

   −(−6)

   ∴ p+p

   2

   =

   4

   3

   ∴ 4p+4p

   2

   =3

   ∴ 4p

   2

   +4p−3=0 ———– ( 1 )

   ⇒ Product of the roots =

   a

   c

   ∴ p×p

   2

   =

   8

   k

   ∴ p

   3

   =

   8

   k

   ——- ( 2 )

   ⇒ 4p

   2

   +4p−3=0 [ From ( 1 ) ]

   ⇒ 4p

   2

   −2p+6p−3=0

   ⇒ 2p(2p−1)+3(2p−1)=0

   ⇒ (2p+3)(2p−1)=0

   ∴ p=

   2

   −3

   and p=

   2

   1

   Now, putting p=

   2

   1

   in equation ( 2 ) we get,

   ⇒ (

   2

   1

   )

   3

   =

   8

   k

   ⇒

   8

   1

   =

   8

   k

   ∴ k=1

   Now, using p=

   2

   −3

   in equation ( 2 )

   ⇒ (

   2

   −3

   )

   3

   =

   8

   k

   ⇒

   8

   −27

   =

   8

   k

   ∴ k=−27

   ∴ Values of k are 1 and −27

   Reply

2. Step-by-step explanation:

   The given quadratic equation is 8x

   2

   −6x+k=0, comparing it with ax

   2

   +bx+c.

   ⇒ Then, a=8,b=−6,c=k

   ⇒ It is given that one root of this equation is square of the other root. So, if we assume one root to be p, the other root can be p

   2

   . So, we assume the roots to be p and p

   2

   .

   ⇒ Sum of the roots =

   a

   −b

   ∴ p+p

   2

   =

   8

   −(−6)

   ∴ p+p

   2

   =

   4

   3

   ∴ 4p+4p

   2

   =3

   ∴ 4p

   2

   +4p−3=0 ———– ( 1 )

   ⇒ Product of the roots =

   a

   c

   ∴ p×p

   2

   =

   8

   k

   ∴ p

   3

   =

   8

   k

   ——- ( 2 )

   ⇒ 4p

   2

   +4p−3=0 [ From ( 1 ) ]

   ⇒ 4p

   2

   −2p+6p−3=0

   ⇒ 2p(2p−1)+3(2p−1)=0

   ⇒ (2p+3)(2p−1)=0

   ∴ p=

   2

   −3

   and p=

   2

   1

   Now, putting p=

   2

   1

   in equation ( 2 ) we get,

   ⇒ (

   2

   1

   )

   3

   =

   8

   k

   ⇒

   8

   1

   =

   8

   k

   ∴ k=1

   Now, using p=

   2

   −3

   in equation ( 2 )

   ⇒ (

   2

   −3

   )

   3

   =

   8

   k

   ⇒

   8

   −27

   =

   8

   k

   ∴ k=−27

   ∴ Values of k are 1 and −27

   Reply