Question

Find the exact values of p for which the equation
px^2+px-1
has real roots

Answer 1

Step-by-step explanation:

Given: px²+px-1

To find: Find the exact values of p for which the equation px²+px-1 has real roots.

Solution:

For a quadratic equation having real roots, must have Discriminant greater or equal to zero.

D ≥ 0

b²-4ac ≥ 0

here,

a = p

b=p

c=-1

${p}^{2} + 4p \geqslant 0 \\ \\ p(p + 4) \geqslant 0 \\ \\ p \neq0$

because for that the equation not longer be a quadratic equation.

$p \geqslant - 4$

Checking for p=-4

$- 4 {x}^{2} - 4x - 1 = 0 \\ \\ or \\ \\ 4 {x}^{2} + 4x + 1 = 0 \\ \\ x_{1,2} = \frac{ - 4 ± \sqrt{ { 4}^{2} - 4(4)(1) } }{8} \\ \\ x_{1,2} = \frac{ - 4 ± 0}{8} \\ \\ x_{1,2} = - \frac{1}{2}$

both roots are real and equal.

For all p which are greater than 0, the equation have real roots.

Final answer:

• For values of p = -4, roots of given quadratic equation are real and equal.
• For all values of p >0,roots of given quadratic equation are real and distinct.

Hope it helps you.