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secondary 3 | A Maths

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I'm not sure of how to do part ii but I know part i. please help me! :-)

Since the roots for b) are 2α + β and α + 2β,

Then (x - (2α + β) )(x - (α + 2β) ) = 0

Sub in the values of α and β directly (or find what are the values of the roots first, then sub in, whichever way you like)

Then expand the equation, and multiply accordingly to get your answer.

If you didn't know how to find α and β,

then find the sum and product of roots.

Sum = 2α + β + α + 2β

= 3α + 3β

= 3(α + β)

Sub the value of α + β you found in i).

Product = (2α + β)(α + 2β)

= 2α² + 4αβ + αβ + 2β²

= 2(α² + 2αβ + β²) + αβ

= 2(α+β)² + αβ

Sub the values of αβ and α + β in i).

You have found the sum and product,

Typical quadratic equation :

ax² + bx + c = 0

Divide by a,

x² + b/a x + c/a = 0

We know sum = -b/a and product = c/a

and thus b/a = -sum

So just substitute the values you found into this equation, and multiply accordingly to get your answer

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