Question

Determine value(s) of p for which the given quadratic equation has real roots x^2+6x+2p+1​

Answer 1

Answer: p should be less than or equal to 4

Step-by-step explanation: For a function to have real roots its determinant must be greater than or equal to 0. As the formula for determinant is

"b^2-4ac" where a,b,c are the coefficients of x^2, x, constant therefore when we solve this further by substituting values from x^2+6x+2p+1 we can say 6^2-4*1*(2p+1) should be greater than or equal to zero

=36-8p-4 is greater than or equal to zero

=  32-8p

=8(4-p)

=p should be less than or equal to 4 (the inequality changed due to multiplication with minus)

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