Question

13. Positive value of p for which equation x2 +pX+64 = 0 and x2 - 8x+ p = 0 vill both have
real roots will be (a) p>16 (b) p<16 (c) p=16 (d) none of these​

Answer 1

Answer:

For the positive value of p=16, both the equations will have real roots.

Step-by-step explanation:

Given data:

Two equation of equalities

x² + px + 64 = 0 ….. (i)

x² + 8x +p = 0 …… (ii)

To find: positive value of p for which both the above equations will have real roots.

We know that, ax²+bx+c=0 will have real roots only when its determinant,

D = (b² - 4ac) ≥ 0.

Equation (i) will have real roots when,

p² – (4 * 1 * 64) ≥ 0

or, p² – 256 ≥ 0

or, p² ≥ 256

or, p ≥ ± 16

∴ p≥16 or p≤(-16) …….. (iii)

Equation (ii) will have real roots when,

8² – (4 * 1 * p) ≥ 0

Or, 64 – 4p ≥ 0

Or, 4p ≥ 64

Or, p ≥ 64/4

∴ p ≥ 16 ………. (iv)  

∴ From equation (iii) & (iv) we can say that, for positive value of p = 16, x in both the equations will give real roots.