Question

Find the value of 'p' if the roots of equation : px² - (2p-2)x+p = 0 has equal real roots


i need answer written in copy plz​

Answer 1

Answer:- p = 1/2

Explanation:-

We are given a polynomial f(x) and this polynomial has equal and real roots. Which is only possible when the discriminant of the given polynomial is 0.

We know,

• D = b² - 4ac

We don't know the values of a, b and c. So let's find it. Comparing the given polynomial with the standard form of quadratic polynomial i.e., ax² + bx + c , we get

• a = p
• b = -(2p - 2)
• c = p

You may notice that the value of b is -(2p - 2) and I have written (2p - 2) in the discriminant formula, This is because the square of either numbers is the same. See here,

⇒ { -(2p - 2) }²

⇒ { -(2p - 2) × -(2p - 2) }

⇒ { (2p - 2) × (2p - 2) }

[ As, product of two negative numbers is always positive. ]

⇒ { (2p - 2) }²

Which means,

⇒ { -(2p - 2) }² = { (2p - 2) }²

This is possible only when the numbers are squared.

Now, After substituting the values in the discriminant formula, we get the value of p as 1/2.

Which is our required answer.

Answer 2

Answer:

Given :-

$\sf \: p {x}^{2} - (2p - 2)x + p = 0$

To Find :-

Value of p

Solution :-

We know that

When roots are equal and real. So,

$\sf{b}^{2} - 4ac = d$

$\sf \: 0 = 0$

So,

By putting value

Here,

b = 2p - 2

a = p

c = p

$\sf \:{ (2p - 2)}^{2} - 4(p)(p) = 0$

$\sf \: {4p}^{2} + 4 - 4 \times 2p - 4{p}^{2}$

$\sf \: 4 {p}^{2} + 4 - 8p - 4 {p}^{2}$

$\sf \: 4 - 8p = 0$

$\sf \: - 8p = 0 - 4$

$\sf \: - 8p = - 4$

$\sf \: 8p = 4$

$\sf \: p = \dfrac{4}{8}$

$\sf \: p \: = \dfrac{1}{2}$