Form an equation with roots which exceed by 2 the roots of the quadr \( 2 x^{2}-(p-4) x-(2 p+1)=0 \)

To find an equation whose roots exceed the roots of \( 2 x^{2}-(p-4) x-(2 p+1)=0 \) by 2, we first need to determine the roots of the original quadratic equation. Using Vieta's formulas, the roots of the quadratic \( ax^2 + bx + c = 0 \) can be expressed as: \[ r_1, r_2 = \frac{-(b/a) \pm \sqrt{(b/a)^2 - 4(c/a)}}{2} \] For our equation, let \( a = 2 \), \( b = -(p-4) \), and \( c = -(2p+1) \). The roots \( r_1 \) and \( r_2\) can be represented based on the coefficients. Now, if \( r_1 \) and \( r_2 \) are the roots of the given equation, the new roots will be \( r_1 + 2 \) and \( r_2 + 2 \). We can construct a new quadratic equation based on its roots by using the fact that for roots \( r_1\) and \( r_2\), the sum and product are given as: 1. Sum: \( (r_1 + 2) + (r_2 + 2) = r_1 + r_2 + 4 \) 2. Product: \( (r_1 + 2)(r_2 + 2) = r_1r_2 + 2(r_1 + r_2) + 4 \) Using Vieta's formulas for the original roots: 1. Sum of roots \( r_1+r_2 = \frac{p-4}{2} \) 2. Product of roots \( r_1r_2 = \frac{-(2p+1)}{2} \) Substituting these values, you can find the new quadratic: \[ x^2 - ((r_1 + r_2) + 4)x + (r_1r_2 + 2(r_1 + r_2) + 4) = 0 \] The final step will give you a new equation in terms of \( p \), which you can simplify to get the desired quadratic form.