Question

2 - 87. complete each of the diamond problems below. the pattern used in the diamond problems is shown at right.

Explanation:

Step1: Identify the diamond - problem rule

In a diamond - problem, the top number is the sum of the two numbers on the sides, and the bottom number is the product of the two numbers on the sides. Let the two side - numbers be \(a\) and \(b\), the top number \(T=a + b\), and the bottom number \(B=a\times b\).

Step2: Solve problem a

Let the two unknown numbers be \(a\) and \(b\). We have \(a + b=8\) and \(a\times b = 8\). From \(a + b=8\), we get \(b = 8 - a\). Substitute into the product equation: \(a(8 - a)=8\), which expands to \(8a-a^{2}=8\), or \(a^{2}-8a + 8=0\). Using the quadratic formula \(a=\frac{-(-8)\pm\sqrt{(-8)^{2}-4\times1\times8}}{2\times1}=\frac{8\pm\sqrt{64 - 32}}{2}=\frac{8\pm\sqrt{32}}{2}=\frac{8\pm4\sqrt{2}}{2}=4\pm2\sqrt{2}\). So the two numbers are \(4 + 2\sqrt{2}\) and \(4 - 2\sqrt{2}\).

Step3: Solve problem b

Let the two numbers be \(a\) and \(b\). We have \(a + b=0\) and \(a\times b=-25\). From \(a + b=0\), we get \(b=-a\). Substitute into the product equation: \(a\times(-a)=-25\), or \(a^{2}=25\), so \(a = 5\) and \(b=-5\) (or vice - versa).

Step4: Solve problem c

Let the two numbers be \(a\) and \(b\). We have \(a + b = 4\) and \(a\times b=-12\). From \(a + b=4\), we get \(b = 4 - a\). Substitute into the product equation: \(a(4 - a)=-12\), which expands to \(4a-a^{2}=-12\), or \(a^{2}-4a - 12=0\). Factoring gives \((a - 6)(a + 2)=0\). So \(a = 6\) and \(b=-2\) (or vice - versa).

Step5: Solve problem d

Let the two numbers be \(a\) and \(b\). We have \(a + b=-2\) and \(a\times b=-15\). From \(a + b=-2\), we get \(b=-2 - a\). Substitute into the product equation: \(a(-2 - a)=-15\), which expands to \(-2a-a^{2}=-15\), or \(a^{2}+2a - 15=0\). Factoring gives \((a + 5)(a - 3)=0\). So \(a=-5\) and \(b = 3\) (or vice - versa).

Answer:

a. \(4 + 2\sqrt{2}\), \(4 - 2\sqrt{2}\)
b. \(5\), \(-5\)
c. \(6\), \(-2\)
d. \(-5\), \(3\)