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Find two numbers whose sum is 55 and whose product is 684

Q: How do you use a quadratic equation to find two real numbers whose sum is 5 and whose product is -14?

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None, if the coefficients of the quadratic are in their lowest form.

When the equation is a polynomial whose highest order (power) is 2. Eg. y= x2 + 2x + 10. Then you can use quadratic formula to solve if factoring is not possible.

If you mean: (x-2)(x+4) = 0 then it is a quadratic equation whose solutions are x = 2 or x = -4

To find the numbers that maximize the product p, we can use the formula for a quadratic equation: x = -b / 2a. Let's call one number x and the other number (120-x). Therefore, the equation becomes x(120 - x^2), which simplifies to -x^3 + 120x. We can find x by setting the derivative equal to zero, resulting in x = 10. Therefore, the two numbers that maximize the product are 10 and 110, with a product of 12100.

Numbers whose product is one is called multiplicative inverses.

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The numbers are 15.75 and -5.75 When tackling probiems like this form a quadratic equation with the information given and solving the equation will give the solutions.

The quadratic formula can be used to find the solutions of a quadratic equation - not a linear or cubic, or non-polynomial equation. The quadratic formula will always provide the solutions to a quadratic equation - whether the solutions are rational, real or complex numbers.

that's impossibleAnswer:xy=9 and x+y=12reduces to the quadratic equation x2-12x+9=0so that once solvedx=-11.196 and y=-0.80385 (approx)

None, if the coefficients of the quadratic are in their lowest form.

2

If you mean: (x-2)(x+4) = 0 then it is a quadratic equation whose solutions are x = 2 or x = -4

When the equation is a polynomial whose highest order (power) is 2. Eg. y= x2 + 2x + 10. Then you can use quadratic formula to solve if factoring is not possible.

2

To find the numbers that maximize the product p, we can use the formula for a quadratic equation: x = -b / 2a. Let's call one number x and the other number (120-x). Therefore, the equation becomes x(120 - x^2), which simplifies to -x^3 + 120x. We can find x by setting the derivative equal to zero, resulting in x = 10. Therefore, the two numbers that maximize the product are 10 and 110, with a product of 12100.

Numbers whose product is one is called multiplicative inverses.

two prime numbers whose product is 141 = 3 & 47

3 and 7 are prime numbers whose product is 21.