Question

இருபடி சமன்பாட்டை தீர்க்க

"(ii) " √2 f^2-6f+3√2=0

Answer 1

Answer:

இருபடி சமன்பாட்டை தீர்க்க

"(ii) " √2 f^2-6f+3√2=0

Explanation:

$45 { - 21}^{2}$

Answer 2

$f=\frac{3+\sqrt{3}}{\sqrt{2}}$  , $f =\frac{3-\sqrt{3}}{\sqrt{2}}$

விளக்கம்:

$\sqrt{2} f^{2}-6 f+3 \sqrt{2}=0$

$a=\sqrt{2}, b=-6, c=3 \sqrt{2}$

$f=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

$=\frac{-(-6) \pm \sqrt{(-6)^{2}-4(\sqrt{2})(3 \sqrt{2})}}{2(\sqrt{2})}$

$=\frac{6 \pm \sqrt{36-12 \sqrt{2} \sqrt{2}}}{2(\sqrt{2})}$

$=\frac{6 \pm \sqrt{36-12(2)}}{2(\sqrt{2})}$

$=\frac{6 \pm \sqrt{36-24}}{2(\sqrt{2})}$

$=\frac{6 \pm \sqrt{12}}{2(\sqrt{2})}$

$=\frac{6 \pm \sqrt{2 \times 2 \times 3}}{2(\sqrt{2})}$

$=\frac{6 \pm \sqrt{3}}{2 \sqrt{2}}$

$f=\frac{6+\sqrt{3}}{2 \sqrt{2}}$

$f=\frac{2(3+\sqrt{3})}{2 \sqrt{2}}$

$f=\frac{3+\sqrt{3}}{\sqrt{2}}$

$f =\frac{6-\sqrt{3}}{2 \sqrt{2}}$

$=\frac{2(3-\sqrt{3})}{2 \sqrt{2}}$

$f =\frac{3-\sqrt{3}}{\sqrt{2}}$

$f=\frac{3+\sqrt{3}}{\sqrt{2}}$  , $f =\frac{3-\sqrt{3}}{\sqrt{2}}$