What are the solutions of $(2x - 1)(x + 3) = 3x + 9$ ?

x = 2

onlyCorrect answerYour answer

x = 2

x = -3

Correct answerYour answer

x = -2

x = 3

Correct answerYour answer

x = \frac{1}{2}

x = -3

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer B

Correct answer: B

Expand the left side: $2x^2 + 5x - 3 = 3x + 9$ .
Bring everything to one side: $2x^2 + 2x - 12 = 0$ , i.e. $x^2 + x - 6 = 0$ .
$(x + 3)(x - 2) = 0$ , so $x = 2$ or $x = -3$ .

Worth noting

The right side is $3(x + 3)$ ; cancelling $(x + 3)$ from both sides loses the root $-3$ and gives 2 only. $\frac{1}{2}$ or $-3$ sets each bracket to zero, which is allowed only when the right side is 0.

Concept: Bringing a quadratic to standard form before factorising
Typical timeTypical: 120 s
Your timeYours: —

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What are the solutions of (2x−1)(x+3)=3x+9(2x - 1)(x + 3) = 3x + 9(2x−1)(x+3)=3x+9?

Explanation

What are the solutions of $(2x - 1)(x + 3) = 3x + 9$ ?