Trying to save more each month, a household carefully tracks the weekly surplus in its food budget, in pounds, using the expression \(3x^2 + 5x - 2\), where...
Question 1 Report
Trying to save more each month, a household carefully tracks the weekly surplus in its food budget, in pounds, using the expression \(3x^2 + 5x - 2\), where \(x\) is the number of weeks since the family started meal-planning.
- Factorise \(3x^2 + 5x - 2\) fully. (2)
- Hence write down the two values of \(x\) for which the surplus is zero. (1)
Answer Details
Factorising a quadratic where the \( x^2 \) coefficient is not 1 means finding two numbers that multiply to give (leading coefficient) \(\times\) (constant term) and add to give the middle coefficient, then splitting the middle term using those numbers.
(a) For \( 3x^2+5x-2 \), the product \( 3 \times (-2) = -6 \) and the required sum is \( 5 \); the numbers \( 6 \) and \( -1 \) satisfy \( 6 \times (-1) = -6 \) and \( 6+(-1)=5 \). Splitting the middle term: \( 3x^2+6x-x-2 = 3x(x+2) -1(x+2) = (3x-1)(x+2) \) [2 marks].
(b) The surplus is zero when either bracket is zero: \( 3x-1=0 \) gives \( x=\dfrac{1}{3} \); \( x+2=0 \) gives \( x=-2 \) [1 mark].
Both roots are mathematically valid answers to "when is the surplus zero", even though \( x=-2 \) (two weeks before meal-planning started) may not make sense in the household's actual timeline; the question only asks for the values, not for a real-world interpretation of each.
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