Questions and Answers #10 Quadratics

Questions and Answers Sheet 10 Quadratics Question #1 How do you solve 64𝑥 2 − 144𝑥 + 81 = 0 ? Answer: We can use Bhaskara here. −𝑏±√𝑏 2 −4𝑎𝑐 2𝑎 where a=64, b=-144 and c=81 144±√144 2 −4(64)(81) 128 144±√20736−20736 128 144±√0 128 144 9 128 = 8 9 You quadratic function is tangent to axis x, touching it only in the point with x coordinate ( ). So, we can 8 that 𝑥1 = 𝑥2 = 9 8 Question #2 How do you find the roots for 𝑥 2 − 14𝑥 − 32 = 0? Answer: In an equation of the following form 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 the method to find the roots is: 1) calculate Δ = 𝑏 2 − 4𝑎𝑐 2) if Δ = 0 there is only one root 𝑥0 = 3) if Δ > 0 there are two roots 𝑥 = −𝑏 2𝑎 −𝑏−√Δ 2𝑎 and 𝑥+ = 4) if Δ < 0 there is no real solution Example: 𝑥 2 − 14𝑥 − 32 = 0 → 𝑎 = 1; 𝑏 = −14; 𝑐 = −32 → Δ = (−14)2 − 4 × 1 × (−32) = 196 + 128 = 324 Δ > 0 therefore we have two roots: 𝑥− = 𝑥+ = 14−√324 2 14+√324 2 = = 14−18 2 14+18 2 4 = − = −2 2 = 32 2 = 16 Let us check the validity of our results: (−2)2 − 14 × (−2) − 32 = 4 + 28 − 32 = 0 OK (16)2 − 14 × (16) − 32 = 256 − 224 − 32 = 0 OK Question #3 How do you find the roots for 𝑓(𝑥) = 𝑥 2 + 12𝑥 + 20? −𝑏+√Δ 2𝑎 Answer: 𝑓(𝑥) = 𝑥 2 + 12𝑥 + 20 Let f(x)=0 0 = 𝑥 2 + 12𝑥 + 20 0 = 𝑥 2 + 2 × 6 × 𝑥 + 62 − 62 + 20 0 = (𝑥 + 6)2 − 36 + 20 = (𝑥 + 6)2 − 16 ∣ +16 16 = (𝑥 + 6)2 ∣ √ ±√16 = 𝑥 + 6 ∣ −6 −6 ± 4 = 𝑥1,2 𝑥1 = −2 or 𝑥2 = −10 Question #4 How do you find the zeroes for 𝑓(𝑥) = 𝑥 2 − 9𝑥 − 70 ? Answer: Step 1 We need to think of two numbers, that, when I add them, sum up to -9, and when I multiply them, have a product of -70. Since the product is negative, we know the signs must be different. Through some thought, we arrive at -14 and 5 as our two numbers, because -14+5=-9 and −14 × 5 = −70 Thus, our equation is as follows: (𝑥 − 14)(𝑥 + 5) = 0 To find the zeroes, we take the opposite signs to get x=14 and x=-5 Question #5 How do you solve 𝑥 2 = 21𝑥? Answer: 𝑥 2 = 21𝑥 rewriting the equation: 𝑥 2 − 21𝑥 = 0 taking out the common term x 𝑥(𝑥 − 21) = 0 equating with 0 x=0 or x-21=0 𝑥 = 0, 21 are the two solutions Question #6 How do you solve 6ℎ2 + 17ℎ + 12 = 0? Answer: 6ℎ2 + 17ℎ + 12 = 0 We can Split the Middle Term of this equation to factorise it and then find the solution for the equation In this technique, if we have to factorise an expression like 𝑎ℎ2 + 𝑏ℎ + 𝑐, e need to think of 2 numbers such that: 𝑁1 × 𝑁2 = 𝑎 × 𝑐 = 6 × 12 = 72 and 𝑁1 + 𝑁2 = 𝑏 = 17 After trying out a few numbers we get 𝑁1 = 9 and 𝑁2 = 8 9 × 8 = 72 , and 9+8=17 6ℎ2 + 17ℎ + 12 = 6ℎ2 + 9ℎ + 8ℎ + 12 = 3ℎ(2ℎ + 3) + 4(2ℎ + 3) = (3ℎ + 4)(2ℎ + 3) the solution for the equation is: 4 ℎ=− ℎ=− 3 3 2 Question #7 How do you solve 3𝑛2 = 10𝑛 ? Answer: (3𝑛)2 = 10𝑛 Dividing both sides by n. (3𝑛)2 𝑛 = 10 𝑛 𝑛 Applying the law of exponents, 3𝑛2−1 = 10 3𝑛 = 10 Dividing 3 by both sides, 3𝑛 10 = 3 3 n=3.33 Question #8 Howe do you solve 𝑥 2 − 12𝑥 + 52 = 0 ? Answer: 𝑥 2 − 12𝑥 + 52 = 0 The equation is of the form 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 where: a=1, b=-12, c=52 The Discriminant is given by: Δ = 𝑏2 − 4 × 𝑎 × 𝑐 = (−12)2 − (4 × 1 × 52) =144-208 =-64 If Δ = 0 then there is only one solution. (for Δ > 0 there are two solutions, for Δ < 0 there are no real solutions) As Δ = −64, this equation has no real solutions Note: The solutions are normally found using the formula 𝑥= −𝑏±√Δ 2×𝑎 As Δ = −64, 𝑥 = −(−12)±√−64 2×1 = 12±√−64 2 Question #9 How do you find its vertex, axis of symmetry, y-intercept and x-intercept for 𝑓(𝑥) = −3𝑥 2 + 3𝑥 − 2 ? Answer: x of vertex =x of axis of symmetry: 𝑏 3 2 −6 𝑥 = (− 𝑎) = − = 1 2 y of vertex: 1 3 3 5 2 4 2 4 𝑦 = 𝑓( ) = − + −2 = − y intercept: y=-2 x-intercept: → 𝑦 = 0 𝐷 = 𝑏 2 − 4𝑐 = 9 − 24 = −15 < 0 There are no real roots (no x-intercepts) because D<0 Since a<0 , the parabola opens downward. The parabola is completely below the x-axis. Question #10 How do you solve 2𝑥 2 + 6𝑥 = 0? Answer: 2𝑥(𝑥 + 3) = 0 equate each factor to zero and solve for x 2𝑥 = 0 ⇒ 𝑥 = 0 𝑥 + 3 = 0 ⇒ 𝑥 = −3