Question

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

Answer:

x²- (1+√2)x + √2 = 0

• a = 1
• b = -(1+√2)
• c = √2

Then, quadratic formula is given by:

$\sf x = \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{2a}$

Substitute the values and solve for x.

$\bullet \ \ \sf x = \dfrac{ -[-(1+\sqrt{2})] \pm \sqrt{(-1-\sqrt{2})^2 - 4(1)(\sqrt{2}) }}{2}$

$\bullet \ \ \sf x = \dfrac{1+\sqrt{2} \pm \sqrt{2-2\sqrt{2}+1}}{2}$

Notice that you can convert radicles into perfect square.

$\bullet \ \ \sf x = \dfrac{1+\sqrt{2} \pm\sqrt{\left(\sqrt{2}\right)^2-2\sqrt{2}\cdot \:1+1^2}}{2}$

The part under square root is in the form a² + b² - 2ab = (a-b)².

$\bullet \ \ \sf x = \dfrac{1+\sqrt{2} \pm \sqrt{\left(\sqrt{2}-1\right)^2 }}{2}$

$\bullet \ \ \sf x = \dfrac{1+\sqrt{2} \pm \left(\sqrt{2}-1\right)}{2}$

Two possibilities exist.

# Case 1:

$\bullet \ \ \sf x = \dfrac{1+\sqrt{2} + \left\sqrt{2}-1\right}{2}$

$\bullet \ \ \sf x = \dfrac{2\sqrt{2} }{2}$

$\bullet \ \ \sf x = \sqrt{2}$

# Case 2:

$\bullet \ \ \sf x = \dfrac{1+\sqrt{2} - \left(\sqrt{2}-1\right)}{2}$

$\bullet \ \ \sf x = \dfrac{1+\sqrt{2} - \sqrt{2}+1}{2}$

$\bullet \ \ \sf x = \dfrac{2}{2}$

$\bullet \ \ \sf x =1$

Therefore, roots are √2 and 1.