ECO 111: INTRODUCTION TO MATHEMATICS FOR ECONOMICS exam. Questions cover roots of quadratic equations, factorization, quadratic formula, sets, community problems, Cramer's rule, and matrices.

Mathematics
ECO 111: INTRODUCTION TO MATHEMATICS FOR ECONOMICS exam. Questions cover roots of quadratic equations, factorization, quadratic formula, sets, community problems, Cramer's rule, and matrices.

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Answer

215-\frac{21}{5}

Step 1: Identify coefficients for the quadratic equation 5x2+21x20=05x^2 + 21x - 20 = 0. The standard form is ax2+bx+c=0ax^2 + bx + c = 0. Here, a=5a=5, b=21b=21, c=20c=-20.

a) Find α+β\alpha + \beta The sum of the roots of a quadratic equation is given by α+β=ba\alpha + \beta = -\frac{b}{a}. α+β=215\alpha + \beta = -\frac{21}{5} The sum of the roots is 215\boxed{-\frac{21}{5}}.

b) Find αβ\alpha \beta The product of the roots of a quadratic equation is given by αβ=ca\alpha \beta = \frac{c}{a}. αβ=205=4\alpha \beta = \frac{-20}{5} = -4 The product of the roots is 4\boxed{-4}.

c) Find α2+β2\alpha^2 + \beta^2 We use the identity α2+β2=(α+β)22αβ\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta. Substitute the values from parts (a) and (b): α2+β2=(215)22(4)\alpha^2 + \beta^2 = \left(-\frac{21}{5}\right)^2 - 2(-4) α2+β2=44125+8\alpha^2 + \beta^2 = \frac{441}{25} + 8 α2+β2=44125+8×2525\alpha^2 + \beta^2 = \frac{441}{25} + \frac{8 \times 25}{25} α2+β2=44125+20025\alpha^2 + \beta^2 = \frac{441}{25} + \frac{200}{25} α2+β2=64125\alpha^2 + \beta^2 = \frac{641}{25} The value of α2+β2\alpha^2 + \beta^2 is 64125\boxed{\frac{641}{25}}.

d) Find α3+β3\alpha^3 + \beta^3 We use the identity α3+β3=(α+β)33αβ(α+β)\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha \beta (\alpha + \beta). Substitute the values from parts (a) and (b): α3+β3=(215)33(4)(215)\alpha^3 + \beta^3 = \left(-\frac{21}{5}\right)^3 - 3(-4)\left(-\frac{21}{5}\right) α3+β3=926112512(215)\alpha^3 + \beta^3 = -\frac{9261}{125} - 12\left(\frac{21}{5}\right) α3+β3=92611252525\alpha^3 + \beta^3 = -\frac{9261}{125} - \frac{252}{5} To combine these, find a common denominator, which is 125: α3+β3=9261125252×25125\alpha^3 + \beta^3 = -\frac{9261}{125} - \frac{252 \times 25}{125} α3+β3=92611256300125\alpha^3 + \beta^3 = -\frac{9261}{125} - \frac{6300}{125} α3+β3=15561125\alpha^3 + \beta^3 = -\frac{15561}{125} The value of α3+β3\alpha^3 + \beta^3 is 15561125\boxed{-\frac{15561}{125}}.

e) Find 1α+1β\frac{1}{\alpha} + \frac{1}{\beta} Combine the fractions: 1α+1β=β+ααβ\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\beta + \alpha}{\alpha \beta} Substitute the values from parts (a) and (b): 1α+1β=2154\frac{1}{\alpha} + \frac{1}{\beta} = \frac{-\frac{21}{5}}{-4} 1α+1β=215×14\frac{1}{\alpha} + \frac{1}{\beta} = \frac{21}{5} \times \frac{1}{4} 1α+1β=2120\frac{1}{\alpha} + \frac{1}{\beta} = \frac{21}{20} The value of 1α+1β\frac{1}{\alpha} + \frac{1}{\beta} is 2120\boxed{\frac{21}{20}}.

f) Find 1α2+1β2\frac{1}{\alpha^2} + \frac{1}{\beta^2} Combine the fractions: 1α2+1β2=β2+α2(αβ)2\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\beta^2 + \alpha^2}{(\alpha \beta)^2} Substitute the values from parts (b) and (c): 1α2+1β2=64125(4)2\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\frac{641}{25}}{(-4)^2} 1α2+1β2=6412516\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\frac{641}{25}}{16}

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Quick Answer

Identify coefficients for the quadratic equation 5x^2 + 21x - 20 = 0. The standard form is ax^2 + bx + c = 0.

ECO 111: INTRODUCTION TO MATHEMATICS FOR ECONOMICS exam. Questions cover roots of quadratic equations, factorization, quadratic formula, sets, community problems, Cramer's rule, and matrices.
Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

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Step 1: Identify coefficients for the quadratic equation 5x^2 + 21x - 20 = 0. The standard form is ax^2 + bx + c = 0. Here, a=5, b=21, c=-20. a) Find + The sum of the roots of a quadratic equation is given by + = -(b)/(a). + = -(21)/(5) The sum of the roots is -(21)/(5). b) Find The product of the roots of a quadratic equation is given by = (c)/(a). = (-20)/(5) = -4 The product of the roots is -4. c) Find ^2 + ^2 We use the identity ^2 + ^2 = ( + )^2 - 2 . Substitute the values from parts (a) and (b): ^2 + ^2 = (-(21)/(5))^2 - 2(-4) ^2 + ^2 = (441)/(25) + 8 ^2 + ^2 = (441)/(25) + (8 × 25)/(25) ^2 + ^2 = (441)/(25) + (200)/(25) ^2 + ^2 = (641)/(25) The value of ^2 + ^2 is (641)/(25). d) Find ^3 + ^3 We use the identity ^3 + ^3 = ( + )^3 - 3 ( + ). Substitute the values from parts (a) and (b): ^3 + ^3 = (-(21)/(5))^3 - 3(-4)(-(21)/(5)) ^3 + ^3 = -(9261)/(125) - 12((21)/(5)) ^3 + ^3 = -(9261)/(125) - (252)/(5) To combine these, find a common denominator, which is 125: ^3 + ^3 = -(9261)/(125) - (252 × 25)/(125) ^3 + ^3 = -(9261)/(125) - (6300)/(125) ^3 + ^3 = -(15561)/(125) The value of ^3 + ^3 is -(15561)/(125). e) Find (1)/() + (1)/() Combine the fractions: (1)/() + (1)/() = ( + )/( ) Substitute the values from parts (a) and (b): (1)/() + (1)/() = (-21)/(5)-4 (1)/() + (1)/() = (21)/(5) × (1)/(4) (1)/() + (1)/() = (21)/(20) The value of (1)/() + (1)/() is (21)/(20). f) Find (1)/(^2) + (1)/(^2) Combine the fractions: (1)/(^2) + (1)/(^2) = (^2 + ^2)/(( )^2) Substitute the values from parts (b) and (c): (1)/(^2) + (1)/(^2) = (641)/(25)(-4)^2 (1)/(^2) + (1)/(^2) = (641)/(25)16 (1)/(^2) + (1)/(^2) = 64125 \