Question

(1) y varies partly as D an partly as the cube root of D.when Y=6, D=27 an when Y=11,D=1/8.find Y when D=3. Correct to 2 decimal point.
(2) using completing the square method find the root of the equation
8x+6x-5=0

Answer 1

Y=k(D)+j(D^1/3)

6=27k + 3j
11=k/8+j/2

multiply second equation by 6, then subtract first equation from second.
6=27k+3j
66=3k/4+5j
or
60=-k(3/4-27)
solve for k
then put that k into either equation, solve for j.

Now solve for y when D=3

Answer 2

1) y = aD + b (cube root of D)
a and b are constant.
when y=6, d=27
6= 27a + b(3√ 27)
6= 27a + 3b ----------equation 1

when y= 11, d = 1/8
11= a/8 + b( 3√ 1/8)
11= a/8 + b/2 --------- equation 2

so we have
6= 27a + 3b
11= a/8 + b/2
Rearrange
27a + 3b =6
a/8 + b/2 =11

Multiply equation 2 by 6 and equation 1 by 1
27a + 3b =6
6a/8 + 6b/2 =66

therefore,
27a + 3b =6
3a/4 + 3b =66

Eliminate 3b in both equation
27a =6
3a/4=66

subtract equ 1 from 2

-105a/4 = 60
Cross Multiply
-105a=240

a= -2.28

Solve for b
6= 27a + 3b
6= 27(-2.28) + 3b
3b=6 - (-61.56)
3b=67.56
b=22.52

Now solve for y when D=3
y = aD + b (cube root of D)
y = (-2.28 x 3) + 22.52 (cube root of 3)
y= -6.84 + 32.48
y= 25.64 to 2 D.P