Question

solve x²-(1+√2)x+√2=0 by quadratic formula​

Answer 1

GIVEN :-

• x² - (1 + √2)x + √2 = 0

TO FIND :-

• Solutions of the given quadratic equation.

SOLUTION :-

As we know that,

$\bigstar \: \boxed{ \displaystyle\sf \: x = \frac{ - b + \sqrt{b ^{2} - 4ac} }{2a} \: , \: x = \frac{ - b - \sqrt{b ^{2} - 4ac } }{2a} }$

$\implies \displaystyle \sf \: x = \frac{ - (1 + \sqrt{2} ) + \sqrt{(1 + \sqrt{2} ) ^{2} - 4 \times 1 \times \sqrt{2} } }{2 \times 1}$

$\implies \displaystyle \sf \: x = \frac{ - (1 + \sqrt{2} ) + \sqrt{1 ^{2} +( \sqrt{2} ) ^{2} + 2 \times \sqrt{2} - 4 \sqrt{2} } }{2}$

$\implies \displaystyle \sf \: x = \frac{ - (1 + \sqrt{2} ) + \sqrt{1 + 2 + \sqrt{2} - 4 \sqrt{2} } }{2}$

$\implies \displaystyle \sf \: x = \frac{ - (1 + \sqrt{2} ) + \sqrt{3 - 2 \sqrt{2} } }{2}$

$\implies \displaystyle \sf \: x = \frac{ - 1 - \sqrt{2} + \sqrt{0.18} }{2}$

$\implies \displaystyle \sf \: x = \frac{ - 1 - \sqrt{2} + 0.42}{2}$

$\implies \displaystyle \sf \: x = \frac{ - 1 - 1.41 + 0.42}{2}$

$\implies \displaystyle \sf \: x = \frac{1.99}{2}$

$\implies \underline{ \boxed{ \displaystyle \sf \: x \approx1}}$

Similarly,

$\implies \displaystyle \sf \: x = \frac{ - (1 + \sqrt{2} ) - \sqrt{(1 + \sqrt{2} ) ^{2} - 4 \times 1 \times \sqrt{2} } }{2 \times 1}$

$\implies \displaystyle \sf \: x = \frac{ - (1 + \sqrt{2} ) - \sqrt{1 ^{2} +( \sqrt{2} ) ^{2} + 2 \times \sqrt{2} - 4 \sqrt{2} } }{2}$

$\implies \displaystyle \sf \: x = \frac{ - (1 + \sqrt{2} ) - \sqrt{1 + 2 + \sqrt{2} - 4 \sqrt{2} } }{2}$

$\implies \displaystyle \sf \: x = \frac{ - (1 + \sqrt{2} ) - \sqrt{3 - 2 \sqrt{2} } }{2}$

$\implies \displaystyle \sf \: x = \frac{ - 1 - \sqrt{2} - \sqrt{0.18} }{2}$

$\implies \displaystyle \sf \: x = \frac{ - 1 - 1.41 - 0.42 }{2}$

$\implies \displaystyle \sf \: x = \frac{2.83}{2}$

$\implies \displaystyle \sf \: x = 1.41$

$\implies \underline{ \boxed{\displaystyle \sf \: x \approx \sqrt{2} }}$