Question

find the positive values of k for which the equation x²+10kx+16=0 has no real roots​

Answer 1

Step-by-step explanation:

let x+1=0

therefore, x=-1

Putting the value of x in the above eq. we get,

(-1)2+10k(-1)+16=0

1+(-10k)+16=0

1-10k+16=0

17-10k=0

-10k=0-17

-10k=-17

therefore, k= 17/10

Answer 2

Given:

A quadratic equation x² + 10kx + 16=0 has no real roots.

To Find:

The positive values of k such that the equation has no real roots are?

Solution:

The given problem can be solved using the concepts of quadratic equations.    

1. The given quadratic equation is x²+10kx+16=0

2. For an equation to have equal roots the value of the discriminant is 0,  

=> The discriminant of a quadratic equation ax² + b x + c = 0 is given by the formula,  

=> Discriminant ( D ) =$\sqrt{b^2-4ac}$ .  

=> For non real roots D < 0.  

3. Substitute the values in the above formula,  

=>  D < 0,  

=> √[(10k)² - 4(1)(16)] < 0,

=> [100k² - 64] < 0,

=> 100k² < 64,

=> k² < 64/100,

=> k < ± (8/10),

=> k < ± (4/5),

4. The positive values of k are k>0 and k< 4/5.

=> k belongs to ( 0, 4/5 ).

Therefore, the positive values of k are k belongs to (0,4/5).