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Question

Find the values of k for which the following equations have real roots

(i) (ii)

(iii) x2-4kx+k=0 (iv) kxx-25+10=0

(v) kx(x-3)+9=0 (vi) 4x2+kx+3=0

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Solution

(i) The given quadric equation is , and roots are real and equal

Then find the value of k.

Here,

As we know that

Putting the value of

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

Therefore, the value of

(ii) The given quadric equation is , and roots are real and equal

Then find the value of k.

Here,

So,

As we know that

Putting the value of

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

Now factorizing of the above equation

So, either

Therefore, the value of

(iii) The given quadratic equation is x2-4kx+k=0, and roots are real and equal.

Then find the value of k.

Here,

x2-4kx+k=0

So,

a=1, b=-4k and c=k.

As we know that D=b2-4ac

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

D=-4k2-41k =16k2-4k

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

So, 16k2-4k=0

Now factorizing the above equation,

16k2-4k=04k4k-1=04k=0 or 4k-1=0k=0 or k=14

Therefore, the value of k=0, 14.

(iv) The given quadratic equation is kxx-25+10=0, and roots are real and equal.

Then find the value of k.

Here,

kxx-25+10=0kx2-25kx+10=0

So,

a=k, b=-25k and c=10.

As we know that D=b2-4ac

Putting the value of a=k, b=-25k and c=10.

D=-25k2-4k10 =20k2-40k

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

So, 20k2-40k=0

Now factorizing the above equation,

20k2-40k=020kk-2=020k=0 or k-2=0k=0 or k=2

Therefore, the value of k=0, 2.

(v) The given quadratic equation is px(x-3)+9=0, and roots are real and equal.

Then find the value of p.

Here,

px(x-3)+9=0px2-3px+9=0

So,

a=p, b=-3p and c=9.

As we know that D=b2-4ac

Putting the value of a=p, b=-3p and c=9.

D=-3p2-4p9 =9p2-36p

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

So, 9p2-36p=0

Now factorizing the above equation,

9p2-36p=09pp-4=09p=0 or p-4=0p=0 or p=4

Therefore, the value of p=0, 4.

(vi) The given quadratic equation is 4x2+px+3=0, and roots are real and equal.

Then find the value of p.

Here,

4x2+px+3=0

So,

a=4, b=p and c=3.

As we know that D=b2-4ac

Putting the value of a=4, b=p and c=3.

D=p2-443 =p2-48

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

So, p2-48=0

Now factorizing the above equation,

p2-48=0p2-432=0p-43p+43=0p-43=0 or p+43=0p=43 or p=-43

Therefore, the value of p=±43.


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