
Plzz Help
1)2x^212x+p=q(xr)^2+10 for all values of x, find the cosntants p,q,r
2)the length of a childrens retangular playgroud is 10m more then the width. The width is x meteres
the perimeter is greater than 64m. Write a linear inequality in x
3)show that (x13)(x+23) < 0
and also what would i need to do to find the equation of a line that passes through the midpoint AB and is perpendicular to AB

For question 1
Here is the concept.
if equation is satisfied by more than 2 values of x then a = b = c =0.
Reduce your equation in standard quadratic form and then equate all coefficients to zero and solve for p, q, r
2) length is x + 10
perimeter 2 ( x + x + 10) = 4x + 20
perimeter is greate than 64.
so 4x + 20 >64
x > 22
3) remaining questions ar not clear.
but still let me tell you that (x13)(x+23) < 0 is true for 23 < x < 13