Question

solve please tell me the end step clearly 2x^2-2√2x+1

Answer 1

☆ ANSWER:-

Given quadratic polynomial:-

$\tt\leadsto2 {x}^{2} - 2 \sqrt{2} x + 1$

To Find:-

• The zeroes of given polynomial.

So,

Let us find out the zeroes by factorization and middle term splitting method.

Therefore, here:-

$\tt\mapsto \: coefficient \: of \: {x}^{2} \times constant \: term. \\ \\ \tt = 2 \times 1 = 2. \\ \\ \tt \: and \\ \\ \tt\mapsto2 = \sqrt{2} \times \sqrt{2} \\ \\ also \\ \\ \tt\mapsto \sqrt{2} + \sqrt{2} = 2 \sqrt{2} = middle \: term.$

So, lets split middle term:-

Also we are finding the zeroes so the polynomial should be equal to zero.

$\tt\mapsto2 {x}^{2} - ( \sqrt{2} + \sqrt{2} ) x+ 1 = 0\\ \\ \tt\mapsto2 {x}^{2} - \sqrt{2} x - \sqrt{2}x + 1 = 0 \\ \\ \tt\mapsto \sqrt{2} x( \sqrt{2}x - 1) - 1( \sqrt{2}x - 1) = 0 \\ \\ \tt\mapsto ( \sqrt{2} x -1 )( \sqrt{2} x - 1) = 0\\ \\ \tt\mapsto \sqrt{2}x - 1 = 0 \\ \\ \tt\mapsto \sqrt{2}x = 1 \\ \\ \tt\mapsto( \sqrt{2}x) {}^{2} = (1) {}^{2} \\ \\ \sf(squaring \: both \: sides) \\ \\ \tt\mapsto2 {x}^{2} = 1 \\ \\ \tt\mapsto {x}^{2} = \frac{1}{2} \\ \\ \tt\mapsto \: x = \frac{ + }{} \sqrt{ \frac{1}{2} } \\ \\ \tt\mapsto \: either \: x = \sqrt{ \frac{1}{2} } \: \: or \: \: x = -\sqrt{ \frac{ 1}{2} }$

$\therefore$ The required roots are $\sqrt{\frac{1}{2}}$ And $-\sqrt{\frac{1}{2}}$.