Question

Find the zeroes of polynomial f(x) = x² + 2√2 - 6 and also verify the relation between zeroes and coefficient.​

Answer 1

Step-by-step explanation:

Given, p(x) = x2 + 2√2x – 6

We put p(x) = 0

⇒ x2 + 2√2x – 6 = 0

⇒ x2 + 3√2x – √2x – 6 = 0

⇒ x(x + 3√2) – √2 (x + 3√2) = 0

⇒ (x – √2)(x + 3√2) = 0

This gives us 2 zeros, for x = √2 and x = -3√2

Hence, the zeros of the quadratic equation are √2 and -3√2.

Now, for verification

Sum of zeros = –coefficient of x/ coefficient of x^2

√2 + (-3√2) = (2√2)/1

-2√2 = -2√2

Product of roots = constant/coefficient of x^2

√2 x (-3√2) = (−6)/2√2

-3 x 2 = -6/1

-6 = -6

Therefore, the relationship between zeros and their coefficients is verified.

Answer 2

$\underline \red{ \pmb{Given : }}$

• F(x) = x² + 2√2 - 6

$\underline \red{ \pmb{To \: find : }}$

• Find zeroes
• Verify the relation between zeroes and coefficient

$\underline \green{ \pmb{Solution : - }}$

$\sf \: \: \: \: \: \qquad \: f(x) = {x}^{2} + 2 \sqrt{2} - 6$

$\sf \: We \: put \: \red{ f(x) = 0}$

$\longrightarrow \sf {x}^{2} + 2 \sqrt{2} - 6 = 0$

$\longrightarrow \sf {x}^{2} + 2 \sqrt{2x} - \sqrt{2x} - 6 = 0$

$\longrightarrow \sf x(x + 3 \sqrt{2} ) - \sqrt{2} (x + 3 \sqrt{2} ) = 0$

$\longrightarrow \sf(x - \sqrt{2} )(x + 3 \sqrt{2} ) = 0$

$\sf \: \pmb {This \: gives \: us \: 2 \: zeroes \: \: , for}\:$

$\qquad \sf \: x = \sqrt{2} \: \: \: and \: \: \: x - 3 \sqrt{2}$

$\sf\orange{\pmb{Hence, \: the \: zeroes \: of \: the \: quadratic \: equation \: are \: \sqrt{2} \: \: and \: \: - 3 \sqrt{2} } }\:$

$\tt \pmb{Now \: for \: verification, }$

$\qquad \underline\red{ \boxed {Sum \: of \: zeros \: = \frac{ - coefficient \: of \: x}{coefficient \: of \: {x}^{2} }} }$

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

$\longrightarrow \: \tt \: - 2 \sqrt{2} = - 2 \sqrt{2}$

$\underline\red{ \boxed{\qquad \: Product \: of \: roots = \frac{constant}{coefficient \: of \: {x}^{2} }} }$

$\longrightarrow \tt \sqrt{2} \times ( - 3 \sqrt{2} ) = \frac{( - 6)}{2 \sqrt{2} }$

$\longrightarrow \tt - 3 \times 2 = \frac{ - 6}{1}$

$\longrightarrow \tt - 6 = - 6$

Therefore, the relationship between zeroes and coefficient is verified