Lesson Find a number using a quadratic equation
Algebra
->
Quadratic Equations and Parabolas
-> Lesson Find a number using a quadratic equation
Log On
Quadratics: solvers
Quadratics
Practice!
Practice
Answers archive
Answers
Lessons
Lessons
Word Problems
Word
In Depth
In
Source code of 'Find a number using a quadratic equation'
This Lesson (Find a number using a quadratic equation)
was created by by
ikleyn(52778)
:
View Source
,
Show
About ikleyn
:
<H2>Find a number using a quadratic equation</H2> <H3>Problem 1</H3>Find two integers with a product of -40 and a sum of -3. <B>Solution</B> <U>ANSWER</U>. The numbers are -8 and 5. <pre> Two solutions are possible. One solution is to guess the answer mentally. A person familiar with the multiplication table, can do it mentally (without using equations) in 10 seconds. Alternative solution is algebraic. Write equations as you read the problem x + y = -3 (1) xy = -40. (2) From equation (1), express y = -3 - x and substitute into equation (2) x*(-3-x) = -40. Simplify it and reduce to the standard form of the quadratic equation -x^2 - 3x + 40 = 0, or, equivalently, x^2 + 3x - 40 = 0. You may solve the last equation either by factoring (x+8)*(x-5) = 0 or using the quadratic formula {{{x[1,2]}}} = {{{(-3 +- sqrt(3^2 - 4*(-40)))/2}}} = {{{(-3 +- sqrt(169))/2}}} = {{{(-3 +- 13)/2}}}. It gives you the same answer that is placed at the very beginning. </pre> <H3>Problem 2</H3>Two positive integers differ by 6. Their product is 616. Find the integers. <B>Solution</B> <pre> Let x be the smaller integer; then the greater integer is (x+6). Then you write this equation for the product x*(x+6) = 616, or x^2 + 6x - 616 = 0. Use the quadratic formula to solve this quadratic equation. {{{x[1,2]}}} = {{{(-6 +- sqrt((-6)^2 + 4*616))/2}}} = {{{(-6 +- sqrt(2500))/2}}} = {{{(-6 +- 50)/2}}} Only positive root is the solution to the problem x = {{{(-6 + 50)/2}}} = {{{44/2}}} = 22. <U>ANSWER</U>. The numbers are 22 and 28. </pre> <H3>Problem 3</H3>The difference of two positive numbers is six. Their product is 223 less than the sum of their squares. What are the two numbers? <B>Solution</B> <pre> Let x be the larger number, y be the lesser number. Then x - y = 6, (1) xy = x^2 + y^2 - 223. (2) Square equation (1) (both sides). Keep equation (2) as is: x^2 - 2xy + y^2 = 36, (3) x^2 - xy + y^2 = 223 (4) (<<<---=== it is transformed eq(2) ) ----------------------------------Subtract eq(3) from eq(4). You will get xy = 187. Now you have system of two equations x - y = 6, xy = 187. It is reduced to the quadratic equation x*(x-6) = 187 x^2 -6x - 187 = 0 {{{x[1,2]}}} = {{{(6 +- sqrt(36 + 4*187))/2}}} = {{{(6 +- 28)/2}}}, {{{x[1]}}} = {{{(6+28)/2}}} = 17, {{{y[1]}}} = 11. {{{x[2]}}} = {{{(6-28)/2}}} = -11, {{{y[2]}}} = -17. <U>Answer</U>. There are TWO solutions: a) (x,y) = (17,11); b) (x,y) = (-11,-17). Since the problem asks about positive numbers, only first pair satisfies this requirement. </pre> <H3>Problem 4</H3>The sum of a number and its reciprocal is 10/3. What is the number? <B>Solution</B> There are two ways to solve this problem. One way is easy and gives the solution in 3-4-5 seconds. The other way is formal, through solving a quadratic equation. I will show you both ways. <U>Easy way</U> <pre> After reading the condition, turn on your mind. {{{10/3}}} = 3 {{{1/3}}} = 3 + {{{1/3}}}. So, the number is 3 and its reciprocal is {{{1/3}}}. It is one answer. Another answer is THIS: The number is {{{1/3}}} and its reciprocal is 3. </pre> Thus easy way is complete. <U>The formal way</U> <pre> Let x be the number; then reciprocal is {{{1/x}}}. The equation is THIS: x + {{{1/x}}} = {{{10/3}}}. Multiply both sides by 3x 3x^2 + 3 = 10x 3x^2 - 10x + 3 = 0 {{{x[1,2]}}} = {{{(10 +- sqrt(10^2 - 4*3*3))/(2*3)}}} = {{{(10 +- sqrt(64))/6}}} = {{{(10 +- 8)/6}}}. One root is {{{x[1]}}} = {{{(10+8)/6}}} = {{{18/6}}} = 3. The other root is {{{x[2]}}} = {{{(10-8)/6}}} = {{{2/6}}} = {{{1/3}}}. Thus we solved the problem formally and mentally (!) </pre> For advanced young students, who are able manipulate fractions freely, it is an entertainment problem, which SHOULD be solved mentally, and it is fan for them to do it. For other students, the formal way is often very boring and torturing way to solve the problem formally. Good student should know both ways. Therefore, I placed both ways solutions here with the easy joking mode first. <H3>Problem 5</H3>The product of two consecutive odd integers is 323. Find the integers. <B>Solution</B> <pre> Let x be the EVEN integer exactly midway between the two odd consecutive integers. Then the odd integers are (x-1) and (x+1), and (x+1)*(x-1) = 323. or x^2 - 1 = 323 x^2 = 323 + 1 = 324 x = +/- {{{sqrt(324)}}} = +/- 18. So, the positive odd integers are 17 and 19; the negative odd integers are -19 and -17. </pre> <H3>Problem 6</H3>A positive real number is 2 less than another. If the sum of the squares of the two numbers is 6, find the numbers. <B>Solution</B> <pre> Let x be the smaller positive real number. Then the larger number is (x+2), and the problem leads to equation x^2 + (x+2)^2 = 6 (the sum of squares is 6). Simplify and find x x^2 + (x^2 + 4x + 4) = 6 2x^2 + 4x -2 = 0 x^2 + 2x - 1 = 0 Use the quadratic formula {{{x[1,2]}}} = {{{((-2) +- sqrt(2^2 - 4*1*(-1)))/2}}} = {{{(-2 +- sqrt(8))/2}}} = {{{-1 +- sqrt(2)}}}. <U>ANSWER</U>. The numbers are x = {{{-1+sqrt(2)}}} (the positive root) and {{{-1-sqrt(2)}}}. </pre> My other lessons on solving quadratic equations in this site are - <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/HOW-TO-solve-quadratic-equation-by-completing-the-square-Learning-by-examples.lesson>HOW TO solve quadratic equation by completing the square - Learning by examples</A> - <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/Solving-quadratic-equations-without-quadratic-formula.lesson>Solving quadratic equations without quadratic formula</A> - <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/Who-is-who-in-quadratic-equations.lesson>Who is who in quadratic equations</A> - <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/Using-Vieta%27s-theorem-to-solve-qudratic-equations.lesson>Using Vieta's theorem to solve quadratic equations and related problems</A> - <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/Using-quadratic-equations-to-solve-word-problems.lesson>Using quadratic equations to solve word problems</A> - <A HREF=https://www.algebra.com/algebra/homework/quadratic/Challenging-word-problems-solved-using-quadratic-equations.lesson>Challenging word problems solved using quadratic equations</A> - <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/Business-related-problems-to-solve-them-using-quadratic-equations.lesson>Business-related problems to solve them using quadratic equations</A> - <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/Rare-beaty-investment-problem-to-solve-using-quadratic-equation.lesson>Rare beauty investment problem to solve using quadratic equation</A> - <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/How-to-solve-the-problem-on-quadratic-equation-mentally-and-avoid-boring-calculations.lesson>HOW TO solve the problem on quadratic equation mentally and avoid boring calculations</A> - <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/Entertainment-problems-on-quadratic-equations.lesson>Entertainment problems on quadratic equations</A> - <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/OVERVIEW-of-lessons-on-solving-quadratic-equations.lesson>OVERVIEW of lessons on solving quadratic equations</A> Use this file/link <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A> to navigate over all topics and lessons of the online textbook ALGEBRA-I.
