Question

Find the set of values of k for which the equation 2x^2-10x+8=kx has no real roots

Answer 1

Step-by-step explanation:

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Answer 2

The values of k is $k<-18,k<-2$

Step-by-step explanation:

Given : Equation $2x^2-10x+8=kx$

To find : The set of values of k for which the equation has no real roots ?

Solution :

Write the equation as,

$2x^2-10x+8=kx$

$2x^2-10x-kx+8=0$

$2x^2-x(10+k)+8=0$

If the roots has no real roots then discriminant is less than 0.

i.e. $b^2-4ac<0$

Here, a=2, b=-(10+k) and c=8

$(-(10+k))^2-4(2)(8)<0$

$100+k^2+20k-64<0$

$k^2+20k+36<0$

$k^2+18k+2k+36<0$

$k(k+18)+2(k+18)<0$

$(k+18)(k+2)<0$

$(k+18)<0,(k+2)<0$

$k<-18,k<-2$

Therefore, the values of k is $k<-18,k<-2$

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