Question

If the equation px^2 + 2x + p= 0 has 2 equal roots, then p=?

Answer 1

The given equation will have equal roots for p = 1 or -1.

Given,

The equation $px^2 + 2x + p= 0$ has 2 equal roots.

To find,

p.

Solution,

A quadratic equation is of the form

$ax^2+bx+c=0$     ...(1)

and, its discriminant is given by,

$D=b^2-4ac$

The discriminant is used to determine the nature of the roots of the quadratic equation, as follows.

If $D > 0,$ the roots of the equation are real and distinct,

If $D= 0,$ the roots of the equation are real and equal,

If $D < 0,$ the roots of the equation are imaginary.

Here, it is given that the roots of the quadratic equation $px^2 + 2x + p= 0$ are equal.

Comparing this equation with (1), we get,

$a=p,\\b=2,\\c=p.$

Now, D for this equation will be,

$D =b^2-4ac=(2)^{2} -4(p)(p)$

$\implies D =4-4p^2$

$\implies D =4(1-p^2)$

As the roots of the equation are given to be equal. So,

$D=0$

$\implies 4(1-p^2)=0$

∵ 4 ≠ 0

∴ $(1-p^2)=0$

$\implies p^2=1$

⇒ p = ± 1.

So, p = 1 or -1.

Therefore, the given equation will have equal roots for p = 1 or -1.

Answer 2

Given  :

1. given equation = $px^{2} +2x+p=0$
2. roots are equal

To find : value of p

solution :

the general form of quadratic equation is

$ax^{2} +bx+c=o ......(1)$

and , the discriminant is given by ,

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

D is used as to determine the nature of roots, we know that ,

• if D = 0 roots are real and equal ,

Here, the root of the equation $px^{2} +2x+p=0$ are equal

by comparing given equation from from (1)

$a = p\\b = 2\\c = p$ ,

now ,

$D=b^{2}-4ac=2^{2}-4(p)(p)\\ = > D=4-4p^{2} \\= > D=4(1-p^{2})$

as the roots are equal, D = 0

$= > 4(1-p^{2})=0\\ = > 4\neq 0\\= > (1-p^{2})=0\\ = > p^{2}=1\\ = > p=+1,-1$

therefore , p = -1 , +1

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