Maths for quadratic Equations

1.
let the smaller number be a
the another number is a+1 a(a+1)=240
a^2+a=240
a^2+a-240=0
(a+16)(a-15)=0
a=-16 (rejected) or a=15the two numbers are 15,162.
let the smaller number be a
the another number is a+4 a^2+(a+4)^2=168
a^2+a^2+8a+16=168
2(a^2)+8a+16-168=0
2(a^2)+8a-152=0
a^2+4a-76=0a={-4+[4^2-4(1)(-76)]^(1/2)} / [2(1)]
or a={-4-[4^2-4(1)(-76)]^(1/2)} / [2(1)]a=-2+4*[5^(1/2)]
or a=-2-4*[5^(1/2)]the two numbers are
( -2+4*[5^(1/2)] and 2+4*[5^(1/2)] )
or ( -2-4*[5^(1/2)] and 2-4*[5^(1/2)] )3.
let the two numbers be a,b
a+b=19
a^2+b^2=265 a=19-b
(19-b)^2+b^2=265
361-38b+b^2+b^2=265
2(b^2)-38b+96=0
b^2-19b+48=0
(b-3)(b-16)=0
b=3 or b=16when b=3,
a=19-3=16when b=16,
a=19-16=3So, the two numbers are 3,164.
If 1+3+5+......+ (2n-1)=289,
n^2=289
n^2-289=0
n^2-17^2=0
(n-17)(n+17)=0
n=17 or n=-17 (rejected) the value of n is 175.
let the base be a cm.
the base=the height*2
a cm=the height*2
the height=a/2 cm the area=base*height
98=a*(a/2)
196=a^2
a^2-196=0
a^2-14^2=0
(a-14)(a+14)=0
a=14 or a=-14 (rejected)So,the base is 14 cm6.
when h=6,
6=5(t^2)+10t+1
5(t^2)+10t-5=0
t^2+2t-1=0 t={-2+[2^2-4(1)(-1)]^(1/2)} / [2(1)]
or t={-2-[2^2-4(1)(-1)]^(1/2)} / [2(1)]t=-1+2^(1/2)
or t=-1-2^(1/2) (rejected)So,after -1+2^(1/2)=0.414213562 (corr. to 9 d.p.) seconds,
the water rocket will be 6 m above the ground