Question

2. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes
and the co-efficients :
${x}^{2} - ( \sqrt{2 } + 1)x + \sqrt{2}$
​

Answer 1

GIVEN :-

• A quadratic polynomial x² - (√2 + 1)x + √2

TO FIND :-

• The zeroes of the quadratic polynomial.

SOLUTION :-

$\\ : \implies \displaystyle \sf \: x ^{2} - \big( \sqrt{2} + 1 \big)x + \sqrt{2} = 0$

$: \implies \displaystyle \sf \: x ^{2} - \sqrt{2} x - x + \sqrt{2} = 0$

$: \implies \displaystyle \sf \: x ^{2} - x - \sqrt{2} x + \sqrt{2} = 0$

$: \implies \displaystyle \sf \: x(x - 1) - \sqrt{2} (x - 1) = 0$

$: \implies \displaystyle \sf \: (x - 1)(x - \sqrt{2} ) = 0$

$: \implies \displaystyle \sf \: x - 1 = 0 \: , \: x - \sqrt{2} = 0$

$: \implies \underline{ \boxed{ \displaystyle \sf \bold{ \: x = 1 \: , \: x = \sqrt{2}} }}$

__________________…

★ Sum of zeros.

$\\: \implies \displaystyle \sf \ \alpha + \beta = \dfrac{- co-efficient \ of \ x}{co-efficient \ of \ x²}$

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

$\\: \implies \displaystyle \sf \ 1 + \sqrt{2} = 1 + \sqrt{2}$

★ Product of zeroes.

$\\: \implies \displaystyle \sf \ \alpha \beta = \dfrac{constant \ term}{co-efficient \ of \ x²}$

$: \implies \displaystyle \sf \ 1 \times \sqrt{2} = \dfrac{\sqrt{ 2}}{1}$

$\\: \implies \displaystyle \sf \ \sqrt{2} = \sqrt{2}$

Answer 2

Given data :-

→ x² - (√2 + 1 )x + √2

Solution :-

Let, α and β are the roots of given quadratic polynomial.

→ x² - (√2 + 1 )x + √2

Now,

→ x² - (1 + √2)x + √2

→ x² - x - √2x + √2

→ x(x - 1) - √2 (x - 1)

→ (x - 1) (x - √2)

→ x - 1 = 0 or x - √2 = 0

→ x = 1 or x = √2

Hence, according to assumption

α = 1 & β = √2

Now, for the relationship between the zeroes and the co-efficients :-

→ x² - (√2 + 1 )x + √2

Compair above equation with

ax² + bx + c = 0

Hence, a = 1, b = (- √2 - 1 ), c = √2

Now, for verification of

→ α + β = -b/a

Here,

→ α + β = 1 + √2. .........( 1 )

→ -b/a = - (- √2 - 1 )/1

→ -b/a = √2 + 1 ..........( 2 )

from eq. ( 1 ) & eq. ( 2 ) we know

α + β = -b/a

Now, for verification of

→ αβ = c/a

Here,

→ αβ = 1 × √2

→ αβ = √2 .........( 3 )

→ c/a = √2/1

→ c/a = √2 .........( 4 )

from eq. ( 3 ) & eq. ( 4 ) we know

αβ = c/a

Hence, the zeroes of the quadratic polynomials is 1 & √2