# What two numbers have a product of 36 and a sum of -13?

## Answers

3 and 12

Step-by-step explanation:

Let x represent one of the numbers and y represent the other number. Then,

Sum: x + y = 15 → y = 15 - x

Product: xy = 36

Use substitution method by replacing "y" with "15 - x" in the second equation, then solve for x:

x(15 - x) = 36

15x - x² = 36

0 = x² - 15x + 36

0 = (x - 3)(x - 12)

0 = x - 3 and 0 = x - 12

x = 3 x = 12

25 i believe

Step-by-step explanation:

3 and 12

3 + 12 = 15

and

3 * 12 = 36

Hope this helps! :)))

3 and 12

Step-by-step explanation:

let the 2 numbers be x and y then

x + y = 15 → (1)

xy = 36 → (2)

From (1) : y = 15 - x → (3)

substitute y = 15 - x into (2)

x(15 - x) = 36

15x - x² = 36 ( subtract 36 from both sides )

15x - x² - 36 = 0 ( multiply through by - 1 and arrange in standard form )

x² - 15x + 36 = 0 ← in standard form

(x - 3)(x - 12) = 0 ← in factored form

equate each factor to zero and solve for x

x - 3 = 0 ⇒ x = 3

x - 12 = 0 ⇒ x = 12

substitute these values into (3) for corresponding value of y

x = 3 → y = 15 - 3 = 12

x = 12 → y = 15 - 12 = 3

Hence the 2 numbers are 3 and 12

The question is asking us to find two numbers which when added will give us 15 and when multiplied will give us 36.

We can find the numbers from the factors of 36 .

Let us look at the numbers that multiply to give us 36.

36=2x13

36=3x123+12=15

36=4x9

36=6x6

36=36x1

Out of these pair of numbers we will look for the pair which when added will give 15

12+3=15.

We got the two numbers which when multiplied will give 36 and when added 15.

The numbers 3 and 12 when multiplied will give 36 and the sum of these numbers is 15.

X - y = 5

x.y = 3.36

[tex](x-y)^2=5^2[/tex]

[tex]x^{2} -2xy + y^{2} =25[/tex]

[tex]x^{2} + y^{2}=25+6.72=31,72[/tex]

[tex](x+y)^2= x^{2} +2xy+ y^{2} =31,72 + 6,72 = 38,44[/tex]

[tex]\sqrt{(x+y)^2}= \sqrt{38,44}[/tex]

x+y= 6,2

and we know that x - y= 5

x+y+x-y=6,2 + 5

2x=11,2

x=5,6

so then y=0,6

x= 5,6

y= 0,6

X-y=5

xy=3.36

add y to both sides on first

x=5+y

sub that in other eqation

(5+y)y=3.36

expand

y^2+5y=3.36

minus 3.36 both sides

y^2+5y-3.36=0

use quadratic formula

for

ay^2+by+c=0

[tex]y= \frac{-b+/- \sqrt{b^2-4ac} }{2a}[/tex]

for 1y^2+5y-3.36

[tex]y= \frac{-5+/- \sqrt{5^2-4(1)(-3.36)} }{2(1)}[/tex]

[tex]y= \frac{-5+/- \sqrt{25+13.44} }{2}[/tex]

[tex]y= \frac{-5+/- \sqrt{38.44} }{2}[/tex]

[tex]y= \frac{-5+/- 6.2}{2}[/tex]

y=-2.5+/-3.1

y=5.6 or 0.6

sub back

x=y+5

so

x=10.6 or 5.6

the numbers are either 10.6 and 5.6 or 0.6 and 5.6

wait,but 10.6 and 5.6 don't multiply to get 3.36 so that is an extrainiesous answer

answer is 0.6 and 5.6

nothing

Step-by-step explanation:

a product of 2 positive numbers always shows a positive value if we add them