Question

Determine the positive value of k for which both the EQ. x² + Kx +64 and x² - 8x + k =0 will have real roots.​

Answer 1

Answer:

Equation : x² + Kx + 64 = 0

= D = k² -256

= D ≥ 0 (for real roots)

and eq. x² - 8x + k =0

D =64 - 4k

D ≥ 0

64 - 4k ≥ 0

k≤ 16

k=16 (using (I) and (ii) EQ. )

thanks ❤️

Answer 2

Solution :-

Formula used :

For the real roots : (b² - 4ac) ≥ 0

x² + Kx + 64

=> b² - 4ac ≥ 0

=> k² - 4 × 1 × 64 ≥ 0

=> k² ≥ 256

=> k² ≥ √256 ≥ 16 _____(i)

x² - 8x + k

=> 0 ≤ b² - 4ac

=> 0 ≤ 8² - 4 × k × 1

=> 4k ≤ 64

=> k ≤ 16 ______(ii)

From equation (i) and (ii)

16 is the positive value of k for which both the EQ.