Question

if a=8+3 root 7,b=1/a,find the value of a2+b2

Answer 1

a = 8+3√7   b = 1/a  ⇒ 1/8+3√7
                                 ⇒ 8-3√7/1  (on rationalising)
 now,    a = 8+3√7    b = 8-3√7
  ∴  a²+b² = (8+3√7)² + (8-3√7)²
               =  (64 + 63+48√7) + (64 + 63 - 48√7)
       a²+b² =   254

Answer 2

The value of $\bf {a}^{2} + {b}^{2}$ is 254.

Given:

• $a = 8 + 3 \sqrt{7}$
• $b = \frac{1}{a}$

To find:

• Find the value of ${a}^{2} + {b}^{2}$

Solution:

Concept/Formula to be used:

Rationalization:

• Any fractional number can be free with any radical sign, by rationalization.
• Rationalization factor of denominator is multiplied with numerator and denominator.
• RF is conjugate of denominator; i.e. if a+√b is denominator then a-√b is RF.

Identity used:

1. $\bf (x - y)(x + y) = {x}^{2} - {y}^{2}$
2. $\bf ( {x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy$
3. $\bf ( {x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy$

Step 1:

As $a = 8 + 3 \sqrt{7}$

thus, rationalization is not required for this, as denominator has no radical sign.

Step 2:

Rationalize the denominator.

$b = \frac{1}{8 + 3 \sqrt{7} }$

Multiply and divide by RF.

$b = \frac{1}{8 + 3 \sqrt{7} } \times \frac{8 - 3 \sqrt{7} }{8 - 3 \sqrt{7} }$

or

use Identity 1 in denominator.

$b = \frac{8 - 3 \sqrt{7} }{( {8)}^{2} - ( {3 \sqrt{7}) }^{2} }$

or

$b = \frac{8 - 3 \sqrt{7} }{64 - 63}$

or

$\bf b = 8 - 3 \sqrt{7 }$

Step 3:

Find square of a and b.

Apply Identity 2.

${a}^{2} = ( {8 + 3 \sqrt{7} )}^{2}$

or

${a}^{2} = ( {8)}^{2} + ( {3 \sqrt{7} )}^{2} + 2 \times 8 \times 3 \sqrt{7}$

or

${a}^{2} = 64 + 63 + 48 \sqrt{7}$

or

$\bf {a}^{2} = 127 +48 \sqrt{7}$

by the same way; apply Identity 3.

$\bf {b}^{2} = 127 - 48 \sqrt{7}$

Step 4:

Add both.

${a}^{2} + {b}^{2} = 127 + 48 \sqrt{7} + 127 - 48 \sqrt{7}$

or

${a}^{2} + {b}^{2} = 254$

Thus,

$\bf \red{{a}^{2} + {b}^{2} = 254 }$

Learn more:

1) Simplify (√5+√2) the whole square

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2) if x = 3 + √8. find the value of x^2 + 1/x^2

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