Answer :
Sure, let's solve the problem step-by-step.
The problem is to find two numbers whose sum is 27 and whose product is 182.
1. Let's denote these two numbers as [tex]\( x \)[/tex] and [tex]\( y \)[/tex].
2. According to the problem, we have two pieces of information:
- Their sum is 27. This gives us the equation:
[tex]\[
x + y = 27
\][/tex]
- Their product is 182. This gives us the equation:
[tex]\[
xy = 182
\][/tex]
3. We now have a system of equations:
[tex]\[
\begin{cases}
x + y = 27 \\
xy = 182
\end{cases}
\][/tex]
4. To find [tex]\( x \)[/tex] and [tex]\( y \)[/tex], we can use these equations. We can solve the first equation for one variable in terms of the other. Let's solve for [tex]\( y \)[/tex]:
[tex]\[
y = 27 - x
\][/tex]
5. Substitute [tex]\( y = 27 - x \)[/tex] into the second equation [tex]\( xy = 182 \)[/tex]:
[tex]\[
x(27 - x) = 182
\][/tex]
6. This expands and simplifies to a quadratic equation:
[tex]\[
27x - x^2 = 182
\][/tex]
[tex]\[
-x^2 + 27x - 182 = 0
\][/tex]
[tex]\[
x^2 - 27x + 182 = 0
\][/tex]
7. To solve this quadratic equation [tex]\( x^2 - 27x + 182 = 0 \)[/tex], we can use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -27 \)[/tex], and [tex]\( c = 182 \)[/tex].
8. Calculate the discriminant:
[tex]\[
\Delta = b^2 - 4ac = (-27)^2 - 4(1)(182) = 729 - 728 = 1
\][/tex]
9. Now plug the values into the quadratic formula:
[tex]\[
x = \frac{27 \pm \sqrt{1}}{2(1)} = \frac{27 \pm 1}{2}
\][/tex]
[tex]\[
x = \frac{27 + 1}{2} \quad \text{or} \quad x = \frac{27 - 1}{2}
\][/tex]
[tex]\[
x = \frac{28}{2} \quad \text{or} \quad x = \frac{26}{2}
\][/tex]
[tex]\[
x = 14 \quad \text{or} \quad x = 13
\][/tex]
10. If [tex]\( x = 14 \)[/tex], then [tex]\( y = 27 - 14 = 13 \)[/tex].
11. If [tex]\( x = 13 \)[/tex], then [tex]\( y = 27 - 13 = 14 \)[/tex].
Thus, the two numbers that satisfy both conditions (sum of 27 and product of 182) are [tex]\( 14 \)[/tex] and [tex]\( 13 \)[/tex].
The problem is to find two numbers whose sum is 27 and whose product is 182.
1. Let's denote these two numbers as [tex]\( x \)[/tex] and [tex]\( y \)[/tex].
2. According to the problem, we have two pieces of information:
- Their sum is 27. This gives us the equation:
[tex]\[
x + y = 27
\][/tex]
- Their product is 182. This gives us the equation:
[tex]\[
xy = 182
\][/tex]
3. We now have a system of equations:
[tex]\[
\begin{cases}
x + y = 27 \\
xy = 182
\end{cases}
\][/tex]
4. To find [tex]\( x \)[/tex] and [tex]\( y \)[/tex], we can use these equations. We can solve the first equation for one variable in terms of the other. Let's solve for [tex]\( y \)[/tex]:
[tex]\[
y = 27 - x
\][/tex]
5. Substitute [tex]\( y = 27 - x \)[/tex] into the second equation [tex]\( xy = 182 \)[/tex]:
[tex]\[
x(27 - x) = 182
\][/tex]
6. This expands and simplifies to a quadratic equation:
[tex]\[
27x - x^2 = 182
\][/tex]
[tex]\[
-x^2 + 27x - 182 = 0
\][/tex]
[tex]\[
x^2 - 27x + 182 = 0
\][/tex]
7. To solve this quadratic equation [tex]\( x^2 - 27x + 182 = 0 \)[/tex], we can use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -27 \)[/tex], and [tex]\( c = 182 \)[/tex].
8. Calculate the discriminant:
[tex]\[
\Delta = b^2 - 4ac = (-27)^2 - 4(1)(182) = 729 - 728 = 1
\][/tex]
9. Now plug the values into the quadratic formula:
[tex]\[
x = \frac{27 \pm \sqrt{1}}{2(1)} = \frac{27 \pm 1}{2}
\][/tex]
[tex]\[
x = \frac{27 + 1}{2} \quad \text{or} \quad x = \frac{27 - 1}{2}
\][/tex]
[tex]\[
x = \frac{28}{2} \quad \text{or} \quad x = \frac{26}{2}
\][/tex]
[tex]\[
x = 14 \quad \text{or} \quad x = 13
\][/tex]
10. If [tex]\( x = 14 \)[/tex], then [tex]\( y = 27 - 14 = 13 \)[/tex].
11. If [tex]\( x = 13 \)[/tex], then [tex]\( y = 27 - 13 = 14 \)[/tex].
Thus, the two numbers that satisfy both conditions (sum of 27 and product of 182) are [tex]\( 14 \)[/tex] and [tex]\( 13 \)[/tex].
