Question

From a quadratic polynamial where zeroes are 3 + root 2 and 3 minus root 2

Answer 1

Question :

From a quadratic polynomial where zeros are 3+√2 and 3-√2.

AnsWer :

x²-6x +7.

Given :

Two roots are given 3+√2 and 3-√2.

To Prove :

Find it's quadratic equation by using two its zeros.

Solution :

Let the zero be,

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt \boxed{\alpha \: \: and \: \: \beta .}$

$\star \: \tt \alpha = 3 + \sqrt{2}. \\ \star\tt \: \beta = 3 - \sqrt{2} .$

so,

Sum of it's Zeros,

$\tt \alpha + \beta = 3 + \cancel{ \sqrt{ 2} }+ 3 - \cancel {\sqrt{2} }. \\ \: \: \: \: \: \: \: \: \: \: \: \: \tt = 6.$

Product of it's Zeros,

$\tt\alpha \times \beta = 3 + \sqrt{2} \times 3 - \sqrt{2} . \\ \tt\red{ \{( a + b)(a - b) = {a}^{2} - {b}^{2} \}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \tt = 9 - 2. \\ \: \: \: \: \: \: \: \: \: \: \: \: \tt= 7.$

Now, We have formula

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{K( {x}^{2} - sx + p)}$

$\star \tt \: \: K \: = constant \: term. \\ \star \tt \: \: S = sum \: of \: zero. \\ \star \tt \: \: P = product \: of \: zero.$

Putting value, in it, we get

$\tt \: \: k( {x}^{2} - 6x + 7).$

Therefore, the quadratic equation whose zeros are 3+√2 and 3-√2 is x²-6x+7.

Let's Verify :

We have equation,

$\: \: \: \tt {x}^{2} - 6x + 7.$

$\tt \: a = 1 \: ,\: b = - 6 \: ,and \: c = 7.$

$\tt \: D = {b}^{2} - 4ac. \\ \: \: \: \: \tt= {( - 6)}^{2} - 4 \times 1 \times 7 .\\</p><p>\: \: \: \: \tt = 36 - 28. \\ \: \: \: \: \tt = 8.$

Now,

$\tt x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a} \\ </p><p>\tt x = \frac{6 \pm \sqrt{8} }{2} \\ \tt \: \: \: = \frac{6 \pm2 \sqrt{2} }{2} \\</p><p> \: \: \: \tt= \frac{ \cancel2(3 \pm \sqrt{2} )}{ \cancel2} \\</p><p>\tt x = 3 \pm \sqrt{2} .$

$\: \: \:\boxed{ \tt \alpha = 3 + \sqrt{2} \: \: and \: \: \beta = 3 - \sqrt{2}}$

$\: \: \: \:\: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \:\: \: \: \: \: \tt\green{Verified}$