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In this lesson, we will learn how to solve complicated equations by making substitutions that reduce the equation to a quadratic.

Q1:

Find the solution set of 5 + 1 2 5 5 = 3 0 π₯ π₯ .

Q2:

Find the solution set of 3 + 7 2 9 3 = 9 0 π₯ π₯ .

Q3:

Factorise fully ( π§ + 3 ) β 1 9 ( π§ + 3 ) + 9 0 2 .

Q4:

Factorise fully ( π₯ β 7 ) + 6 ( π₯ β 7 ) β 7 2 2 .

Q5:

Factorise fully ( π¦ β 4 ) β 1 8 ( π¦ β 4 ) + 8 0 2 .

Q6:

Factorise fully 2 4 π₯ β 3 4 π₯ + 1 2 π₯ 5 3 .

Q7:

Consider the equation π₯ + 2 = 2 π₯ . Adding π₯ + π₯ ο¨ to both sides produces

Is the solution set of the equation formed the same as that of the original equation?

Q8:

Find the solution set of π₯ β 3 3 π₯ + 3 2 = 0 5 5 2 .

Q9:

Find the solution set of π₯ β 7 3 0 π₯ + 7 2 9 = 0 1 2 5 6 5 .

Q10:

Find the solution set of π₯ β 9 π₯ + 1 4 = 0 2 3 1 3 .

Q11:

Find the solution set of 2 β 2 0 Γ 2 + 2 5 6 = 0 2 π₯ + 2 π₯ + 2 .

Q12:

Find the solution set of 2 + 8 = 6 Γ 2 2 π₯ π₯ .

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