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The sum of a number and five times its reciprocal is 4.5. Write a quadratic equation for the number, and use the quadratic formula to find two such numbers. Explain your steps.

2 Answers

3 votes

Explanation:

Let the required number be x

therefore it's reciprocal =
1 over x

According to the given condition:


x + 5 * (1)/(x) = 4.5 \\ \\ x + (5)/(x) = (45)/(10) \\ \\\frac{ {x}^(2) + 5}{x} = (9)/(2) \\ \\ 2( {x}^(2) + 5) = 9x \\ 2 {x}^(2) + 10 - 9x = 0 \\ \red{ \boxed {\bold{2{x}^(2) - 9x + 5 = 0 }}}\\ is \: the \: required \: quadratic \: equation. \\

User The Guest
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7 votes

Answer:

x= -2 and (x+3)=0, x= -3

Explanation:

"The sum of a number" well we do not know what the number is right, so let's call the number x. So we know we have x and we are adding it to something because it says sum. What are we adding it to? "5 times its reciprocal". A reciprocal of a number is equal to 1/the number. So the reciprocal of x is just 1/x. But we want 5 times this so multiply by 5. So now we have x + 5(1/x). Now we need to set it equal to something to be able to solve it. The problem says the sum of a number and 5 times its reciprocal is -6, so x + 5(1/x) = -6. Now just use order of operations to solve for x. First multiply: x + 5/x =-6. Then the next stperfect is harder because we want to get x by itself but we have x + 5/x. A trick you can do here is multiply the whole equation by x. This cancels out the fraction for us: x(x + 5/x) = -6(x) becomes x^2 +5 = -6x. Now we have a quadratic equation, to solve this you have to Factor or use the quadratic formula. I'll try factoring: first we need all of the terms on one side: x^2 +5x +6 =0. Now we need factors of 6 that add or subtract to 5. There are a couple of options here we could use 6 and -1 or 2 and 3. 6 and -1 would not work because both signs in the equation are plus which means we need our factors to both be positive, also if you tried 6 and -1 and then foiled out the factors to check it, it would not work. So 2 and 3 have to be the factors so the equations equals (x+2)(x+3)=0. Now we set each factor equal to zero and solve for x: (x+2)=0, x= -2 and (x+3)=0, x= -3. Those are the two numbers that would work! Hope this helps!!

User Sergei Rybalkin
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