Question

find the roots of the following quadratic equation 2x2- 2 root 2x + 1 = 0

Answer 1

2x² - 2√2x + 1 = 0

2x² - √2x - √2x + 1 = 0

√2x ( √2x - 1 ) - 1 ( √2x - 1 ) = 0

( √2x - 1 ) ( √2x - 1 ) = 0


* ( √2x - 1 ) = 0
x = 1/√2

* ( √2x - 1 ) = 0
x = 1/√2

Answer 2

Answer:

Step-by-step explanation:

Concept:

A quadratic equation is one with a degree of two, indicating that the function's largest exponent is two.

A quadratic has the usual form $y = ax^{2} + bx + c$

If a, b, and c are all numbers, and a can't be zero.

Given:

The quadratic equation is $2 x^{2}-2 \sqrt{2} x+1=0$.

Find:

Find the roots for the given equation.

Solution:

Consider the following quadratic equation is $2 x^{2}-2 \sqrt{2} x+1=0$.

When it comes to quadratic equations, it's worth comparing it to the conventional form $\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}=0$, we get $a=2, b=-2 \sqrt{2}, c=1$

As a result, the quadratic equation's roots are, $\mathrm{x}=\frac{-\mathrm{b} \pm \sqrt{\mathrm{b}^{2}-4 \mathrm{ac}}}{2 \mathrm{a}}$

$=\frac{-(-2 \sqrt{2}) \pm \sqrt{(-2 \sqrt{2})^{2}-4 \times 2 \times 1}}{2 \times 2}$

$\begin{aligned}&=\frac{(2 \sqrt{2}) \pm \sqrt{8-8}}{4} \\&=\frac{2 \sqrt{2} \pm 0}{4} \\&=\frac{2 \sqrt{2}}{4}\end{aligned}$

$\begin{aligned}&=\frac{\sqrt{2}}{2} \\&=\frac{1}{\sqrt{2}}\end{aligned}$

Therefore, The roots are $\frac{1}{\sqrt{2} } , \frac{1}{\sqrt{2} }$.

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