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Academic Skills Advice Solving Quadratics A quadratic equation is one where the highest power is 2. There are 3 ways of solving them: factorising, using the formula and completing the square: This lesson will look at the method of factorisation. (See the previous lesson for how to use the formula or complete the square.) Factorising when the coefficient of ππ is 1: Factorising a quadratic usually means putting it into 2 brackets (i.e. the reverse of multiplying out brackets seen in lesson 3). For example: Factorises to become: ππ + ππ β ππ (π + π)(π β π) This is the same expression factorised into 2 brackets. )(π₯ To factorise an expression, write 2 brackets with an π₯ in each [(π₯ )] then follow the systematic steps to work out the correct signs and numbers to go in the brackets: ο· 1st look at the sign of the number on its own: +ππ number means the signs in the brackets will be the same: (π₯ + )(π₯+ ) or (π₯ β )(π₯β ) βππ number means that the signs in the brackets will be different: (π₯ + )(π₯β ) Then follow the correct route below: If Same signs: ο· ο· e.g.(π₯ + )(π₯+ ) or (π₯ β )(π₯β ) Look at the coefficient of π: If itβs +ve then the signs in the brackets are +ve if itβs βve then the signs in the brackets are βve (π₯ + )(π₯+ ) (π₯ β )(π₯β ) Finally find the numbers for each bracket: The 2 numbers need to multiply to give the number on its own, and add to give the coefficient of π₯. If Different signs: e.g. (π₯ + )(π₯β ) ο· Find the numbers for each bracket: The 2 numbers need to multiply to give the number on its own, and have a difference of the coefficient of π₯. ο· Look at the coefficient of π: If itβs +ve then put the biggest number in the bracket with the + sign. if itβs βve then put the biggest number in the bracket with the β sign. © H Jackson 2012 / ACADEMIC SKILLS 1 Examples: ο· Factorise: ππ + ππ + ππ Step 1: Step 2: Step 3: Step 4: we have +ππ so the signs must be the same. we have +ππ so the signs must both be positive. so far we have, (π + )(π+ ) work out the numbers (3x4=12 and 3+4=7) Answer: (π₯ + 4)(π₯ + 3) ο· Factorise: ππ β ππ + ππ Step 1: Step 2: Step 3: Step 4: we have +ππ so the signs must be the same. we have βππ so the signs must both be negative. so far we have, (π β )(πβ ) work out the numbers (4x5=20 and 4+5=9) Answer: (π₯ β 4)(π₯ β 5) ο· Factorise: ππ + ππ β ππ Step 1: Step 2: Step 3: Step 4: we have βππ so the signs must be the different. so far we have, (π + )(πβ ) work out the numbers (8x5=40 and 8-5=3) we have +ππ so the biggest number goes with the + sign. Answer: (π₯ + 8)(π₯ β 3) ο· Factorise: ππ β ππ β ππ Step 1: Step 2: Step 3: Step 4: we have βππ so the signs must be the different. so far we have, (π + )(πβ ) work out the numbers (6x3=18 and 6-3=3) we have βππ so the biggest number goes with the - sign. Answer: (π₯ + 3)(π₯ β 6) (n.b. once you have the 2 brackets you can multiply them out to check that you get what you started with.) © H Jackson 2012 / ACADEMIC SKILLS 2 Factorising when the coefficient of ππ is not 1: If the coefficient of π₯ 2 is not 1 you can still use the above method but you would have to use trial and error to decide what the coefficients of π₯ should be in the brackets. The following shows a systematic method for factorising these types of quadratics: Examples: ο· Factorise π ππ β πππ β ππ π ππ β πππ β ππ Step 1: do 6 × β10 = β60 Step 2: find 2 numbers that multiply to β60 and add or subtract to β11. Step 3: ±1 × β60 = β60, canβt make β11 X ±2 × β30 = β60, canβt make β11 X ±3 × β20 = β60, canβt make β11 X ±4 × β15 = β60, β15 + 4 = β11 οΌ (±5 × β12 = β60, canβt make β11 X) (±6 × β10 = β60, canβt make β11 X) Split the middle value (using the numbers you found) and rewrite the equation: (Notice this is the same equation written differently: β15π₯ + 4π₯ = β11π₯) π ππ β πππ + ππ β ππ Step 4: Split the equation into 2 and factorise each half: π ππ β πππ + ππ β ππ 3π₯(2π₯ β 5) 2(2π₯ β 5) (n.b. the bit in the bracket should always be exactly the same for each half) Step 5: Write the 2 brackets: (3π₯ + 2)(2π₯ β 5) (Notice the (3π₯ + 2) comes from the bits outside the brackets) ο· Factorise: π ππ + πππ + π Step 3: 8 × 3 = 24 find 2 numbers that multiply to 24 and add or subtract to 10. 6 × 4 = 24, 6 + 4 = 10 οΌ (n.b. if thereβs nothing else to 2 8 π₯ + 6π₯ + 4π₯ + 3 Step 4: 2π₯(4π₯ + 3) Step 5: Write the 2 brackets: (2π₯ + 1)(4π₯ + 3) = 0 Step 1: Step 2: + 1(4π₯ + 3) factorise you need to put a 1 outside the bracket.) (n.b. if you prefer just to learn one method, the above can also be used when the coefficient of π₯ 2 =1) © H Jackson 2012 / ACADEMIC SKILLS 3 Factorising can be used to solve a quadratic (i.e. find the values of π₯ that make it equal zero). The values that you find are called the βrootsβ of the equation β if you sketched the graph these points would be where the graph crosses the π₯-axis. Solving quadratics: The quadratic must = 0 (if it doesnβt then rearrange it). Remember: If 2 things multiply together to give 0, then at least one of them must =0. We use this fact to solve a quadratic once we have factorised it. Letβs look at the 6 examples that we already factorised, put them =0 and solve them. Examples: ο· Solve the following by factorising: Factorise to give: ππ + ππ + ππ = π (π₯ + 4)(π₯ + 3) = 0 (Bracket x bracket = 0, so one of the brackets must be 0.) Either: Rearrange: (π₯ + 4) = 0 or: π₯ = β4 or: ο· ππ β ππ + ππ = π Solve the following by factorising: Factorise to give: Either: (π₯ + 3) = 0 π₯ = β3 (π₯ β 4)(π₯ β 5) = 0 (π₯ β 4) = 0 or: (π₯ β 5) = 0 β΄ π₯ = 4 ππ π₯ = 5 ο· Solve the following by factorising: Factorise to give: Either: ππ + ππ β ππ = π (π₯ + 8)(π₯ β 5) = 0 (π₯ + 8) = 0 or: (π₯ β 5) = 0 β΄ π₯ = β8 ππ π₯ = 5 ο· Solve the following by factorising: Factorise to give: Either: ππ β ππ β ππ = π (π₯ + 3)(π₯ β 6) = 0 (π₯ + 3) = 0 or: (π₯ β 6) = 0 β΄ π₯ = β3 ππ π₯ = 6 © H Jackson 2012 / ACADEMIC SKILLS 4 ο· Solve the following by factorising: π ππ β πππ β ππ = π Factorise to give: Either: (3π₯ + 2)(2π₯ β 5) = 0 (3π₯ + 2) = 0 or: (2π₯ β 5) = 0 2 5 3 2 β΄ π₯ = β ππ π₯ = ο· Solve the following by factorising: π ππ + πππ + π = π Factorise to give: Either: (2π₯ + 1)(4π₯ + 3) = 0 (2π₯ + 1) = 0 or: (4π₯ + 3) = 0 1 3 2 4 β΄ π₯ = β ππ π₯ = β One more thing: ο· Solve: βππ β ππ + ππ = π This looks a bit more tricky because we have βπ₯ 2 . Probably the easiest way to solve this is to multiply everything by β1 (i.e. change all the signs) and then you can factorise as normal. Now we have: ππ + ππ β ππ = π (try it and check that your answers work) (we donβt need to put β0 as itβs the same thing.) © H Jackson 2012 / ACADEMIC SKILLS 5