WMA11 October 2022

4. WMA11/01 Edexcel IAL P1 October 2022, Q4 (Quadratics)

In this question you must show detailed reasoning.

Solutions relying on calculator technology are not acceptable.

f(x) = x2 (2x + 1) – 15x 

(a) Solve f(x) = 0​​ 

(4)

(b) Hence solve y​​ 

y43 (2y23+ 1) – 15y23= 0            y > 0

giving your answer in simplified surd form.​​ 

(2)

SOLUTION

a-​​ 

x22x+1-15x=0

2x3+x2-15x=0

x2x2+x-15=0

2x2+x-15=0

Using breaking the middle term or factorization method to solve the quadratic equation.​​ 

2x2+6sx-5x-15=0

2xx+3-5 x+3=0

x+3 2x-5= 0

x=-3,        x=52 

b-​​ 

On comparing the given equation to​​ x2 2x+1-15x=0, we may conclude that​​ 

y23=x

So equating​​ y23​​ one by one with the soilution of x found in above part.​​ 

When​​ x=0

y23=0

y=0

When​​ x=52

y23=52

y2332=5232

y=523

y=1258

Rationalising the denominator.​​ 

You may take help of the following flash card to learn about rationalising denominators.​​ 

y=5522 x22

y=5104 only 

y=5104

When​​ x=-3

y23= -3

y2332=-332 

y=-33 

y=-27     (Rejected)

Hence, the values of y are​​ 

y=0 & 5104