Question

A Solve the equation 2y2 – 7y – 4 = 0 . b Use your answer to part a to solve the equation 2(2x)2 _ 7(2x) _ 4 = 0. [5] 9

Answer 1

Answer:

$(a)\ y = 4\ or\ y = -\frac{1}{2}$

$(b)\ x = 2 \ or\ x = -\frac{1}{4}$

Step-by-step explanation:

Solving (a)

$2y^2 - 7y - 4 = 0$

Expand

$2y^2 +y - 8y - 4 = 0$

Expand

$y(2y + 1) -4(2y + 1) = 0$

Factor out 2y + 1

$(y -4)(2y + 1) = 0$

Split

$y - 4 =0\ or\ 2y + 1 = 0$

Solve:

$y = 4\ or\ y = -\frac{1}{2}$

Solving (b):

$2(2x)^2 - 7(2x) - 4 = 0$

Compare the above with the equation in (a)

$2y^2 - 7y - 4 = 0$

We have:

$2x = y$

So:

$2x = 4 \ or\ 2x = -\frac{1}{2}$

Solve for x in both cases

$x = 2 \ or\ x = -\frac{1}{4}$