Math, asked by mdmanan4321, 1 year ago

If the roots of quadratic equatron kx²-5x+k=0 are real and equal, the value of kis
(a)≠ 5 (b)≠5/2 (c)≠2/5 (d)≠2​

Answers

Answered by gsrmurthyae
0

Answer:

not equal to 5by 3

Step-by-step explanation:

great thanks very much appreciated if the product and 5-k the best way is new edition by Alex King of

Answered by Rudra0936
6

Answer:

  • Given a Quadratic equation
  • kx^{2}  - 5x + k = 0

A/Q ,we have to find the value of the K for which the given Quadratic have real and equal roots ✓

So, we can determine its value I,e the value of K by using the determinant formula which is given by

b^{2}  -4ac

But there is a condition that the value of the will be such for that the Given Quadratic equation have roots which is real and equal

_____________________________________________

So the for the real and equal roots

b^{2} - 4ac = 0

I,e discriminent is equal to 0

Let us not find the value of K which is as follows✓

 =  > b^{2}  - 4ac = 0 \\  here \: a = k \:  \:  \: b =  - 5 \:  \:  \: and \: c = k

now we will put the values of a,b, and c in the equation so as to get the value of K

 =  > ( - 5)^{2}  - 4 \times k \times k = 0 \\  \\  =  > 25 - 4k^{2}  = 0 \\  \\  =  >  - 4k ^{2}  =  - 25 \\  \\  =  > 4k ^{2}  = 25 \\  \\  > k ^{2}  =  \frac{24}{4}  \\  \\  =  > k =  \sqrt{ \frac{25}{4} }  \\  \\  =  > k =  \frac{5}{2}

the \: value \: \: of \:the \: k \: is \:  \frac{5}{2}

Similar questions