Question

hey guys plz solve this question.....​

Answer 1

Explanation:

To solve :

$\sf{2x^{2}-8x+4~~using~quadratic~formula}$

Solution :

Quadratic formula :-

$\underline{ \boxed{ \sf{ \pmb{ \frac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a} }}}}$

　

a = 2

b = –8

c = 4

$\sf{Discriminant,~D=b^{2}-4ac}$

$\sf\implies{(-8)^{2}-4(2)(4)}$

$\sf\implies{64-4(8)}$

$\sf\implies{64-32}$

$\sf{~~~~~ \therefore~{\purple{\underline{\boxed{\sf{\pmb{Discriminant,~b^{2}-4ac=32}}}}}}}$

$\sf\implies{ {\dfrac{-(-8) \pm {\sqrt{32}}}{2(2)}}}$

$\sf\implies{ {\dfrac{8 \pm {\sqrt{16 \times 2}}}{4}}}$

$\sf\implies{ {\dfrac{8 \pm 4{\sqrt{2}}}{4}}}$

$\sf\implies{ {\dfrac{8 + 4{\sqrt{2}}}{4}}~,~~{\dfrac{8 - 4{\sqrt{2}}}{4}}}$

$\sf\implies{ {\purple{\underline{\boxed{\sf{\pmb{2+4{\sqrt{2}}~,~~2-4{\sqrt{2}}}}}}}}}$

∴ Required answer :

Roots are :

$\sf{{\purple{\pmb{2+4{\sqrt{2}}}}}~~and~~{\purple{\pmb{2-4{\sqrt{2}}}}}}$