Question

Find the value of k for which the quadratic equation kx^2-5x+k=0 have real roots

Answer 1

Answer:

Step-by-step explanation:

here a=k , b=-5 ,c=k

i.e. the roots are real so D  (discriminant)=0

so,b^2-4ac=0

 (-5)^2-4(k)(k)=0

25-4k^2=0

25=4k^2

k^2=25/4

taking root both sides

k=5/2  ANSWER

Answer 2

Answer:

$\huge\bold\star\red{Answer}\star$

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Given:-

• $k {x}^{2} - 5x + k = 0 \: has \: real \: roots$

To find:-

• value of k

Solution:-

If an equation has real roots,then D=0

from the given eq,

$D=0$

$=>b^2-4ac=0$

$=>(-5)^2-4.k.k=0$

$=>25-4k^2=0$

$=>(-4k^2)=(-25)$

$=>k^2=25/4$

$=>k^2=(5/2)^2$

$=>k=5/2$

•°•required value of k is 5/2

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$\huge\green{Hope\:this\:help\:you}$