Question

find a quadratic polynomial whose zeroes are (5-3√2) and (5+3√2)​

Answer 1

Given that,

One zero = $(5 - 3 \sqrt{2})$

Other zero = $(5 + 3 \sqrt{2})$

To find,

A quadratic equation = ?

Assumption:

Let one zero be $\alpha$ and other zero be $\beta$.

Here, we can easily find a quadratic equation from the two zeros, just by using the following formula:

$\boxed{\bold{\mathsf{k[x^2+(\alpha + \beta)x + (\alpha \times \beta)]}}}$

Where k is any constant term.

By substituting the given values,

$\bold{\mathsf{k[x^2+(5-3\sqrt{2}) + (5 + 3 \sqrt{2})]x + [(5-3\sqrt{2}) \times (5+3\sqrt{2})]}}$

$\bold{\mathsf{k[x^2+(10)]x + (7)}}$

$\boxed{\bold{\mathsf{k[x^2+(10)] x + (7)}}}$

Substituting k = 1 (for cancelling k)

$\bold{\mathsf{1[x^2+(10)]x + 7}}$

$\boxed{\bold{\mathsf{x^2 + 10x + 7}}}$

Therefore, the required quadratic equation is x² + 10x + 7.

Answer 2

Answer:

Hello ! Your answer is x²-10x+7

Step-by-step explanation:

    x = (5-3√2)  ,  x = (5+3√2)

⇒ (x-5+3√2) (x-5-3√2)                              [Transposing the sides]

= [x-5+(3√2)] [(x-5-(3√2)]                         {Taking (x-5) as a & (3√2) as b}

= (x-5)² - (3√2)²                                         [∵Using (a+b) (a-b) = a²-b²]

= x²-10x+25 - 18

= x²-10x+7

∴ x²-10x+7 can be a quadratic polynomial.

                                         Hope, it will help you........