Solve the quadratic equation \(6x^{2}+11x-10=0\) by factorising.
Question 1 Report
Solve the quadratic equation \(6x^{2}+11x-10=0\) by factorising.
Answer Details
This quadratic has a leading coefficient of 6, so factorising needs a pair of numbers that multiply to \(6\times(-10)=-60\) and add to \(+11\). Those numbers are \(+15\) and \(-4\).
- Split the middle term: \(6x^{2}+15x-4x-10=0\).
- Factorise in pairs: \(3x(2x+5)-2(2x+5)=0\).
- Take out the common bracket: \((3x-2)(2x+5)=0\) [M1]. Correct use of the quadratic formula also earns this mark.
A product equals zero only when one of its factors is zero, so each bracket is set to zero in turn.
- \(3x-2=0\) gives \(x=\dfrac{2}{3}\) [A1].
- \(2x+5=0\) gives \(x=-\dfrac{5}{2}\) [A1] oe.
Check the factorisation by expanding: \(6x^{2}+15x-4x-10=6x^{2}+11x-10\), which matches the original equation.
The mistake to avoid is reading the roots straight off the brackets as 2 and \(-5\). Each bracket must actually be solved, because the coefficients of \(x\) are not 1.
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