WMA11 June 2021

2. WMA11/01 (Edexcel) IAL P1 June 2021 Q2 Quadratic & Exponential Equations

In this question you must show all stages of your working.

Solutions relying on calculator technology are not acceptable.

f(x) = ax3 + 6a + 8x2 – a2x

where a is a positive constant.​​ 

Given f(–1) = 32​​ 

(a)

(i)​​ show that the only possible value for a is 3

(ii)​​ Using a = 3 solve the equation f(x) = 0​​ 

(5)

(b) Hence find all real solutions of​​ 

(i)​​ 3y+26y23-9y13

(ii)​​ 3(93z) + 2692z – 99z = 0​​ 

(5)

SOLUTION

i-​​ 

It is given that​​ 

f-1=32

a-13+6a+8-12-a2-1=32

-a+6a+8+a2=32

a2+5a-24=0

Solving the quadratic equation.​​ 

a+8a-3=0

a+8=0    &        a-3=0

a=-8     &      a=3

Since ‘a’ is a positive constant,​​ a≠-8. Hence,​​ 

a=3 

ii-​​ 

fx=ax3+6a+8x2-a2x

Substituting the value of ‘a’​​ as 3.

3x3+26x2-9x=0

x3x2+26x-9=0

x3x-1x+9=0

x=0          3x-1=0         x+9=0

                     x=13         x=-9 

Hence, the values of x are​​ 

x=0 ,13, -9

b-i-​​ 

3y+26y23-9y13

Let x = y1/3

So the equation becomes​​ 

3x3+26x2-9x=0

The roots of the cubic equation has already been found in previuos part. So,​​ 

x=0,  x=13,      x=-9

Substituting back x = y1/3

y13=0,     y13=13,  y13=-9

y=0, y=127, y=729

b-ii-​​ 

3(93z) + 2692z – 99z = 0

Let x = 9z

3x3+ 26x2-9x=0

The roots of the cubic equation has already been found in previuos part. So,​​ 

x=0,  x=13,      x=-9

Substituting back x = 9z

9z=0,        9z=13,  9z= -9

      No Solution          9z=13            No Solution         

9z=13

Making the bases same so that the indices rule may be applied.

32z= 3-1

32z= 3-1

2z=-1

z=-12