If `sqrt(86-60sqrt(2)) = a - bsqrt2` then what will be the value of `sqrt(a^(2) + b^(2))`, correct to one decimal place ?

Options

A. 7.2
B. 8.2
C. 8.4
D. 7.8

Answer

To solve the equation \( \sqrt{86 - 60\sqrt{2}} = a - b\sqrt{2} \), we will follow these steps:

### Step 1: Square both sides
Start by squaring both sides of the equation to eliminate the square root:

\[ 86 - 60\sqrt{2} = (a - b\sqrt{2})^2 \]

### Step 2: Expand the right side
Now, expand the right side using the formula \( (x - y)^2 = x^2 - 2xy + y^2 \):

\[ (a - b\sqrt{2})^2 = a^2 - 2ab\sqrt{2} + 2b^2 \]

### Step 3: Set the equation
Now, we can set the equation from Step 1:

\[ 86 - 60\sqrt{2} = a^2 + 2b^2 - 2ab\sqrt{2} \]

### Step 4: Compare coefficients
Now, we can compare the coefficients of the rational and irrational parts:

1. For the rational part:
\[ a^2 + 2b^2 = 86 \]

2. For the irrational part:
\[ -2ab = -60 \implies 2ab = 60 \implies ab = 30 \]

### Step 5: Solve the system of equations
Now we have a system of equations:
1. \( a^2 + 2b^2 = 86 \)
2. \( ab = 30 \)

From the second equation, we can express \( a \) in terms of \( b \):
\[ a = \frac{30}{b} \]

Substituting this into the first equation:

\[ \left(\frac{30}{b}\right)^2 + 2b^2 = 86 \]

### Step 6: Simplify the equation
This gives us:

\[ \frac{900}{b^2} + 2b^2 = 86 \]

Multiply through by \( b^2 \) to eliminate the fraction:

\[ 900 + 2b^4 = 86b^2 \]

Rearranging gives us a polynomial equation:

\[ 2b^4 - 86b^2 + 900 = 0 \]

### Step 7: Let \( x = b^2 \)
Let \( x = b^2 \). Then the equation becomes:

\[ 2x^2 - 86x + 900 = 0 \]

### Step 8: Use the quadratic formula
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[ x = \frac{86 \pm \sqrt{(-86)^2 - 4 \cdot 2 \cdot 900}}{2 \cdot 2} \]

Calculating the discriminant:

\[ x = \frac{86 \pm \sqrt{7396 - 7200}}{4} = \frac{86 \pm \sqrt{196}}{4} = \frac{86 \pm 14}{4} \]

### Step 9: Solve for \( x \)
This gives us two possible values for \( x \):

1. \( x = \frac{100}{4} = 25 \)
2. \( x = \frac{72}{4} = 18 \)

### Step 10: Find \( b \) and \( a \)
Taking the square root gives us:

1. \( b^2 = 25 \implies b = 5 \)
2. \( b^2 = 18 \implies b = \sqrt{18} = 3\sqrt{2} \)

Now using \( ab = 30 \):

1. If \( b = 5 \), then \( a = \frac{30}{5} = 6 \).
2. If \( b = 3\sqrt{2} \), then \( a = \frac{30}{3\sqrt{2}} = 5\sqrt{2} \).

### Step 11: Calculate \( \sqrt{a^2 + b^2} \)
Now, we calculate \( \sqrt{a^2 + b^2} \):

1. For \( a = 6 \) and \( b = 5 \):
\[ \sqrt{6^2 + 5^2} = \sqrt{36 + 25} = \sqrt{61} \]

2. For \( a = 5\sqrt{2} \) and \( b = 3\sqrt{2} \):
\[ \sqrt{(5\sqrt{2})^2 + (3\sqrt{2})^2} = \sqrt{50 + 18} = \sqrt{68} \]

### Step 12: Final calculation
Calculating \( \sqrt{61} \) and \( \sqrt{68} \):

1. \( \sqrt{61} \approx 7.81 \)
2. \( \sqrt{68} \approx 8.25 \)

### Conclusion
The final answer, correct to one decimal place, is:

\[ \sqrt{61} \approx 7.8 \]