Respuesta :
For the first problem, the integers are 8 and 11.
The second problem has missing information. So does the third. They cannot be worked without the information that's missing.
Explanation:
Let x be the larger integer and y be the smaller integer. We know that the smaller integer, y, is 3 less than the larger integer, x, so:
y=x-3
The sum of their squares is 185. This means:
x²+y²=185
x²+(x-3)²=185
x²+(x-3)(x-3)=185
Multiplying the binomials we get:
x²+x*x-3*x-3*x-3(-3)=185
x²+x²-3x-3x+9=185
Combining like terms we get:
2x²-6x+9=185
When solving quadratics, we want the function to equal 0; subtract 185 from both sides:
2x²-6x+9-185=185-185
2x²-6x-176=0
Use the quadratic formula:
[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ \\=\frac{--6\pm \sqrt{(-6)^2-4(2)(-176)}}{2(2)}=\frac{6\pm \sqrt{36--1408}}{4} \\ \\=\frac{6\pm \sqrt{1444}}{4}=\frac{6\pm 38}{4}=\frac{6-38}{4}\text{ or }\frac{6+38}{4}=\frac{-32}{4}\text{ or }\frac{44}{4}=-8\text{ or }11[/tex]
Since both integers must be positive, x = 11. This means that y=11-3=8.
The second problem has missing information. So does the third. They cannot be worked without the information that's missing.
Explanation:
Let x be the larger integer and y be the smaller integer. We know that the smaller integer, y, is 3 less than the larger integer, x, so:
y=x-3
The sum of their squares is 185. This means:
x²+y²=185
x²+(x-3)²=185
x²+(x-3)(x-3)=185
Multiplying the binomials we get:
x²+x*x-3*x-3*x-3(-3)=185
x²+x²-3x-3x+9=185
Combining like terms we get:
2x²-6x+9=185
When solving quadratics, we want the function to equal 0; subtract 185 from both sides:
2x²-6x+9-185=185-185
2x²-6x-176=0
Use the quadratic formula:
[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ \\=\frac{--6\pm \sqrt{(-6)^2-4(2)(-176)}}{2(2)}=\frac{6\pm \sqrt{36--1408}}{4} \\ \\=\frac{6\pm \sqrt{1444}}{4}=\frac{6\pm 38}{4}=\frac{6-38}{4}\text{ or }\frac{6+38}{4}=\frac{-32}{4}\text{ or }\frac{44}{4}=-8\text{ or }11[/tex]
Since both integers must be positive, x = 11. This means that y=11-3=8.
