Question

the roots of the equation x^2+(2p-1)x+p^2=0 are real if​

Answer 1

Answer:

b^2-4ac=0

Step-by-step explanation:

(2p-1)^2-4.1.1=0

Answer 2

Step-by-step explanation:

An equation has real roots if the discriminant is greater than or equal to zero.

for the equation

$a {x}^{2} + bx + c = 0$

the Discriminant is given by

${b}^{2} - 4ac$

we need to solve for the equation

${b}^{2} - 4ac \geqslant 0$

comparing our equation with the stand equation , we get

$a = 1$

$b = 2p - 1$

$c = {p}^{2}$

substitute in the equation and simplify like this

${(2p - 1)}^{2} - 4 {p}^{2} \geqslant 0 \\ 4 {p}^{2} - 4p + 1 - {4p}^{2} \geqslant 0 \\ 1 \geqslant 4p \\ p \leqslant \frac{1}{4}$