Question

px²-6x-2 In this question determine the set of values of p for which quadratic eqation has real roots :​

Answer 1

Answer :-

→ P ≤ 9/2

To Find :-

→ Set of values of P .

Explanation :-

We have ,

$\:$

→ Px² - 6x -2 = 0

Given that it has real roots .

We know that if we have a quadratic equation

→ ax² + bx + c = 0 and it's roots are real so it's discriminate (D) be greater then or equal to zero

D = b² - 4ac ≥0

Compare both of the equation

→ a = P , b = -6 and c = -2

Now ,its discriminate (D) ≥0,

→ (-6)² - 4×P ×(-2) ≥0

→ 36 + 8P≥ 0

→ 8P ≥ -36

→ P ≥ -36/8

→ P ≥ -9/2

Or

→ P≤ 9/2

Answer 2

$\tt{\huge{\orange {------------ }}} S.D$

QUESTION :

px²-6x-2 In this question determine the set of values of p for which quadratic eqation has real roots :

SOLUTION :

Given Polynomial => px^2 - 6x - 2

Given Condition : The polynomial has real roots .

=> This is possible when the value of the discriminant let the quadratic equation is greater than of equal to Zero.

=> D >_ 0

We know that the discriminant of any quadratic Polynomial is equal to B^2 - 4AC where A, B, C are the coefficients of the Polynomial..

=> { -6 }^2 - 4p× { -2} >_ 0

=> 8p > _-36

=> p >_ { -9 / 2 }

So when the value of P is greater than or equal to { -9 / 2} then the Equation has real roots..