Question

Find the roots , if they exist, by applying the quadratic formula or SHRIDHARACHARYA'S RULE​

Answer 1

GIVEN :-

• A quadratic polynomial = 2x² - 2√2 x + 1 = 0

TO FIND :-

• The roots of the given quadratic polynomial by quadratic Formula.

SOLUTION :-

Let recall what is quadratic formula,

★ x = [-b ± √(b² - 4ac)]/2a

• a → 2
• b → 2√2
• c → 1

→ x = [-b ± √(b² - 4ac)]/2a

→ x = [ -(-2√2) ± √{(-2√2)² - 4 × 2 × 1}]/2 × 2

→ x = [2√2 ± √{(-2√2)(-2√2) - 4 × 2 × 1}]/4

→ x = [2√2 ± √{(4√4) - 8 × 1}]/4

→ x = 2√2 ± √{(4 × 2) - 8}/4

→ x = 2√2 ± √8-8/4

→ x = 2√2 ± √0)/4

→ x = 2√2/4

→ x = √2/2

As we know that the descriminant is 0. Hence the quadratic polynomial 2x² - 2√2 x + 1 = 0 has real and equal roots.

Hence the roots of quadratic polynomial 2x² - 2√2 x + 1 = 0 are √2/2 and √2/2.

Answer 2

GIVEN:

• The eq. is 2x² - 2√2x + 1 = 0

FIND:

• roots of the quadratic eq.

SOLUTION:

we know that,

$\sf x = \frac{ - b± \sqrt{ {b}^{2} - 4ac } }{2a}$

where,

a = 2, b = -2√2 and c = 1

substitute these values in the formula

$\sf \longrightarrow x = \frac{ - (-2 \sqrt{2} )± \sqrt{ {( - 2 \sqrt{2} )}^{2} - 4(2)(1) } }{2(2)}$

$\sf \longrightarrow x = \frac{ \not - ( \not-2 \sqrt{2} )± \sqrt{ 8 - 4(2) } }{2(2)}$

$\sf \longrightarrow x = \frac{ 2 \sqrt{2} ± \sqrt{ 8 - 8 } }{4}$

$\sf \longrightarrow x = \frac{ 2 \sqrt{2} ± \sqrt{ 0} }{4}$

$\sf \longrightarrow x = \frac{ 2 \sqrt{2} ± 0 }{4}$

Now,

$\sf x_{1} = \frac{2 \sqrt{2} + 0}{4} = \frac{2 \sqrt{2} }{4}$

$\sf x_{2} = \frac{2 \sqrt{2} - 0}{4} = \frac{2 \sqrt{2} }{4}$

$\sf Hence, the \: roots \: of \: quadratic \: eq. \: 2 {x}^{2} - 2 \sqrt{2} x + 1 = 0 \\ \sf are \boxed{ \sf \frac{2 \sqrt{2} }{4} \: and \: \frac{2 \sqrt{2} }{4} }$