For what value of k the quadratic equation 2 2 40/25 0 KXX − has equal roots also find the roots?

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The given quadric equation is 2kx2 - 40x + 25 = 0, and roots are real and equal

Then find the value of k.

Here, a = 2k, b = -40 and c = 25

As we know that D = b2 - 4ac

Putting the value of a = 2k, b = -40 and c = 25

= (-40)2 - 4 x (2k) x (25)

=1600 - 200k

The given equation will have real and equal roots, if D = 0

Thus,

1600 - 200k = 0

200k = 1600

k = 1600/200

k = 8

Therefore, the value of k = 8.

Page 2

The given quadric equation is 9x2 - 24x + k = 0, and roots are real and equal

Then find the value of k.

Here, a = 9, b = -24 and c = k

As we know that D = b2 - 4ac

Putting the value of a = 9, b = -24 and c = k

= (-24)2 - 4 x (9) x (k)

= 576 - 36k

The given equation will have real and equal roots, if D = 0

Thus,

576 - 36k = 0

36k = 576

k = 576/36

k = 16

Therefore, the value of k = 16.

Page 3

The given quadric equation is 4x2 - 3kx + 1 = 0, and roots are real and equal

Then find the value of k.

Here, a = 4, b = -3k and c = 1

As we know that D = b2 - 4ac

Putting the value of a = 4, b = -3k and c = 1

= (-3k)2 - 4 x (4) x (1)

= 9k2 - 16

The given equation will have real and equal roots, if D = 0

Thus,

9k2 - 16 = 0

9k2 = 16

k2 = 16/9

`k=sqrt(16/9)`

`k=+-4/3`

Therefore, the value of `k=+-4/3` .

Page 4

The given quadric equation is x2 - 2(5 + 2k)x + 3(7 + 10k) = 0, and roots are real and equal

Then find the value of k.

Here, a = 1, b = -2(5 + 2k) and c = 3(7 + 10k)

As we know that D = b2 - 4ac

Putting the value of a = 1, b = -2(5 + 2k) and c = 3(7 + 10k)

= (-2(5 + 2k))2 - 4 x (1) x 3(7 + 10k)

= 4(25 + 20k + 4k2) - 12(7 + 10k)

= 100 + 80k + 16k2 - 84 - 120k

= 16 - 40k + 16k2

The given equation will have real and equal roots, if D = 0

Thus,

16 - 40k + 16k2 = 0

8(2k2 - 5k + 2) = 0

2k2 - 5k + 2 = 0

Now factorizing of the above equation

2k2 - 5k + 2 = 0

2k2 - 4k - k + 2 = 0

2k(k - 2) - 1(k - 2) = 0

(k - 2)(2k - 1) = 0

So, either

k - 2 = 0

k = 2

Or

2k - 1 = 0

2k = 1

k = 1/2

Therefore, the value of k = 2, 1/2.

Page 5

The given quadric equation is (3k+1)x2 + 2(k + 1)x + k = 0, and roots are real and equal

Then find the value of k.

Here, a = (3k + 1), b = 2(k + 1) and c = k

As we know that D = b2 - 4ac

Putting the value of a = (3k + 1), b = 2(k + 1) and c = k

= (2(k + 1))2 - 4 x (3k + 1) x (k)

= 4(k2 + 2k + 1) - 4k(3k + 1)

= 4k2 + 8k + 4 - 12k2 - 4k

= -8k2 + 4k + 4

The given equation will have real and equal roots, if D = 0

Thus,

-8k2 + 4k + 4 = 0

-4(2k2 - k - 1) = 0

2k2 - k - 1 = 0

Now factorizing of the above equation

2k2 - k - 1 = 0

2k2 - 2k + k - 1 = 0

2k(k - 1) + 1(k - 1) = 0

(k - 1)(2k + 1) = 0

So, either

k - 1 = 0

k = 1

Or

2k + 1 = 0

2k = -1

k = -1/2

Therefore, the value of k = 1, -1/2

Page 6

The given quadric equation is kx2 + kx + 1 = -4x2 - x, and roots are real and equal

Then find the value of k.

Here,

kx2 + kx + 1 = -4x2 - x

4x2 + kx2 + kx + x + 1 = 0

(4 + k)x2 + (k + 1)x + 1 = 0

So,

a = (4 + k), b = (k + 1) and c = 1

As we know that D = b2 - 4ac

Putting the value of a = (4 + k), b = (k + 1) and c = 1

= (k + 1)2 - 4 x (4 + k) x (1)

= (k2 + 2k + 1) - 16 - 4k

= k2 - 2k - 15

The given equation will have real and equal roots, if D = 0

Thus,

k2 - 2k - 15 = 0

Now factorizing of the above equation

k2 - 2k - 15 = 0

k2 - 5k + 3k - 15 = 0

k(k - 5) + 3(k - 5) = 0

(k - 5)(k + 3) = 0

So, either

k - 5 = 0

k = 5

Or

k + 3 = 0

k = -3

Therefore, the value of k = 5, -3.

Page 7

The given quadric equation is (k + 1)x2 + 2(k + 3)x + (k + 8) = 0, and roots are real and equal

Then find the value of k.

Here,

a = (k + 1), b = 2(k + 3) and c = (k + 8)

As we know that D = b2 - 4ac

Putting the value of a = (k + 1), b = 2(k + 3) and c = (k + 8)

= (2(k + 3))2 - 4 x (k + 1) x (k + 8)

= (4k2 + 24k + 36) - 4(k2 + 9k + 8)

= 4k2 + 24k + 36 - 4k2 - 36k - 32

= -12k + 4

The given equation will have real and equal roots, if D = 0

-12k + 4 = 0

12k = 4

k = 4/12

k = 1/3

Therefore, the value of k = 1/3.

Page 8

The given quadric equation is x2 - 2kx + 7k - 12 = 0, and roots are real and equal

Then find the value of k.

Here,

a = 1, b = -2k and c = 7k - 12

As we know that D = b2 - 4ac

Putting the value of a = 1, b = -2k and c = 7k - 12

= (-2k)2 - 4 x (1) x (7k - 12)

= 4k2 - 28k + 48

The given equation will have real and equal roots, if D = 0

4k2 - 28k + 48 = 0

4(k2 - 7k + 12) = 0

k2 - 7k + 12 = 0

Now factorizing of the above equation

k2 - 7k + 12 = 0

k2 - 4k - 3k + 12 = 0

k(k - 4) - 3(k - 4) = 0

(k - 3)(k - 4) = 0

So, either

k - 3 = 0

k = 3

Or

k - 4 = 0

k = 4

Therefore, the value of k = 4, 3.

Page 9

The given quadric equation is (k + 1)x2 - 2(3k + 1)x + 8k + 1 = 0, and roots are real and equal

Then find the value of k.

Here,

a = k + 1, b = -2(3k + 1)x and c = 8k + 1

As we know that D = b2 - 4ac

Putting the value of a = k + 1, b = -2(3k + 1)x and c = 8k + 1

= (-2(3k + 1))2 - 4 x (k + 1) x (8k + 1)

= 4(9k2 + 6k + 1) - 4(8k2 + 9k + 1)

= 36k2 + 24k + 4 - 32k2 - 36k - 4

= 4k2 - 12k

The given equation will have real and equal roots, if D = 0

4k2 - 12k = 0

k2 - 3k = 0

Now factorizing of the above equation

k(k - 3) = 0

So, either

k = 0

Or

k - 3 = 0

k = 3

Therefore, the value of k = 0, 3.

Page 10

The given quadric equation is (2k + 1)x2 + 2(k + 3)x + k + 5 = 0, and roots are real and equal

Then find the value of k.

Here,

a = (2k + 1), b = 2(k + 3) and c = k + 5

As we know that D = b2 - 4ac

Putting the value of a = (2k + 1), b = 2(k + 3) and c = k + 5

={2(k + 3)}2 - 4 x (2k + 1) x (k + 5)

= {4(k2 + 6k + 9)} - 4(2k2 + 11k + 5)

= 4k2 + 24k + 36 - 8k2 - 44k - 20

= -4k2 - 20k + 16

The given equation will have real and equal roots, if D = 0

-4k2 - 20k + 16 = 0

-4(k2 + 5k - 4) = 0

k2 + 5k - 4 = 0

Now factorizing the above equation

k2 + 5k - 4

`k=(-b+-sqrt(b^2-4ac))/(2a)`

`k=(-5+-sqrt(25+16))/2`

`k=-5+-sqrt41/2`

So, either

Therefore, the value of `k=-5+-sqrt41/2`

Page 11

The given quadric equation is 4x2 - 2(k + 1)x + (k + 4) = 0, and roots are real and equal

Then find the value of k.

Here,

a = 4, b = -2(k + 1) and c = k + 4

As we know that D = b2 - 4ac

Putting the value of a = 4, b = -2(k + 1) and c = k + 4

= {-2(k + 1)}2 - 4 x 4 x (k + 4)

= {4(k2 + 2k + 1)} - 16(k + 4)

= 4k2 + 8k + 4 - 16k - 64

= 4k2 - 8k - 60

The given equation will have real and equal roots, if D = 0

4k2 - 8k - 60 = 0

4(k2 - 2k - 15) = 0

k2 - 2k - 15 = 0

Now factorizing of the above equation

k2 - 2k - 15 = 0

k2 + 3k - 5k - 15 = 0

k(k + 3) - 5(k + 3) = 0

(k + 3)(k - 5) = 0

So, either

k + 3 = 0

k = -3

Or

k - 5 = 0

k = 5

Therefore, the value of k = -3, 5.

Page 12

The given equation is \[4 x^2 - 2(k + 1)x + (k + 1) = 0\] where a = 4, b = -2(k+1), c = (k+1)
          As we know that D = b2 - 4ac
          Putting the value of a = 4, b = -2(k+1), c = (k+1)

\[\left\{ - 2(k + 1) \right\}^2 - 4 \times 4 \times (K + 1)\]

\[4(K + 1 )^2 - 16(K + 1)\]

\[(K + 1)\left\{ 4(K + 1) - 16 \right\}\]

\[(K + 1)(4K - 12)\]

\[4(K + 1)(K - 3)\]

         For real and equal roots D = 0

\[4\left( K + 1 \right)\left( K - 3 \right) = 0\]

\[K = - 1 \text { or } k = 3\]

    Therefore, the value of