Question

if root of quadratic equation px^2+6x+1=0 are real then find p​

Answer 1

Answer:

answer is 9

pic gives explanation

Answer 2

The value of p ≤ 9

Given :

The quadratic equation px² + 6x + 1 = 0

To find :

The value of p if roots of the quadratic equation are real

Solution :

Step 1 of 3 :

Write down the given equation

Here the given Quadratic equation is

px² + 6x + 1 = 0

Step 2 of 3 :

Find discriminant of the quadratic equation

The given equation is px² + 6x + 1 = 0

Discriminant of the quadratic equation

= b² - 4ac

= 6² - 4 × p × 1

= 36 - 4p

Step 3 of 3 :

Find the value of p

Here it is given that the roots of the quadratic equation are real

∴ Discriminant of the quadratic equation ≥ 0

$\displaystyle \sf{ \implies 36 - 4p \geqslant 0}$

$\displaystyle \sf{ \implies - 4p \geqslant - 36}$

$\displaystyle \sf{ \implies p \leqslant 9}$

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