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Question

(i) Find the values of k for which the quadratic equation 3k+1x2+2k+1x+1=0 has equal roots. Also, find the roots.

(ii) Write all the values of k for which the quadratic equation x2 + kx + 16 = 0 has equal roots. Find the roots of the equation so obtained.

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Solution

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

Then, find the value of k.

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

As we know that D=b2-4ac

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

D=2k+12-43k+11 =4(k2+2k+1)-12k-4 =4k2+8k+4-12k-4 =4k2-4k

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

Thus, 4k2-4k=0
4k(k-1)=0k=0 or k-1=0k=0 or k=1

Therefore, the value of k is 0 or 1.

Now, for k = 0, the equation becomes

x2+2x+1=0x2+x+x+1=0x(x+1)+1(x+1)=0(x+1)2=0x=-1, -1

for k = 1, the equation becomes

4x2+4x+1=04x2+2x+2x+1=02x(2x+1)+1(2x+1)=0(2x+1)2=0x=-12, -12

Hence, the roots of the equation are -1 and -12.

(ii) x2 + kx + 16 = 0

It is given that the quadratic equation has equal roots.
Therefore, Discriminant is equal to zero.

x2+kx+16=0Discriminant=0b2-4ac=0k2-4116=0k2-64=0k2=64k=±8

Hence, the values of k is ±8.

Now,
For k = 8,
The equation becomes
x2 + 8x + 16 = 0
⇒ (x + 4)2 = 0
⇒ x = −4


For k = −8,
The equation becomes
x2 − 8x + 16 = 0
⇒ (x − 4)2 = 0
⇒ x = 4

Hence, the roots of the equation so obtained are 4 and −4.

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