Question

Determine the positive values of P for which equation x²+2px+64=0 and x²-8x+2p=0 will both have real roots

Answer 1

For an equation to have real roots, its determinant should be greater than or equal to 0.

using above stated rule on 1st equation

4*p²- 4*64 >=0

p²>=64

p>=8 or p<=-8

Equation x²-8x+2p

64-8p>=0

8>=p

Answer is ( -∞ , -8) U {8}

Answer 2

Answer: The answer is p = 8.

Step-by-step explanation:  We know that a quadratic equation $ax^2+bx+c=0~(a\neq 0)$ will have real roots if the discriminant D is greater than or equal to 0.

The discriminant is given by

$D=b^2-4ac.$

For the equation, $x^2+2px+64=0$ to have real roots, we have

$(2p)^2-4\times 1\times 64\geq 0\\\\\Rightarrow 4p^2-256\geq 0\\\\\Rightarrow p^2\geq 64\\\\\Rightarrow p\leq -8,~~p\geq 8.~~~~~~~~~~~~~~~~~~~(A)$

And for the equation $x^2-8x+2p=0$ to have real roots, we must have

$(-8)^2-4\times 1\times 2p\geq 0\\\\\Rightarrow 64-8p\geq 0\\\\\Rightarrow p\leq 8.~~~~~~~~~~~~~~~~~~(B)$

Comparing equations (A) and (B), we have

the equations will have real roots only if p = 8.

Thus, the positive value of p is 8.