Question

Find the value of p for which the quadratic equation x(x-4)+p=0 has real roots.


Plzzz solve this

Answer 1

Real roots means b2-4ac=0
So putting the value from the equation then,
(-4)^2-4(1)(p)=0
16-4p=0
16=4p
4=p or
p=4

Answer 2

Given:

A quadratic equation x (x - 4) + p = 0.

To find:

The values of p for which the given quadratic equation has real roots.

Solution:

The value of p for which the given quadratic equation has real roots is greater than or equal to 4.

To answer this question, we will follow the following steps:

As we know in a quadratic equation,

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

the a and b are coefficients of x2 and x respectively.

Also, if the above quadratic equation has real roots then

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

Now,

As given, we have a quadratic equation,

x (x - 4) + p = 0

This can be written as

${x}^{2} - 4x + p = 0$

Where

a = 1, b = -4 and c = p

So,

For real roots,

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

$16 - 4p \geqslant 0$

$4p \geqslant 16$

$p \geqslant 4$

Hence, for real roots, the value of p should be greater than or equal to 4.