Solve the polynomial. y=2x^{2}-4x+1

Question:

Solve the polynomial.

{eq}y = 2x^{2} - 4x + 1 {/eq}

Quadratic Polynomial; Solutions:

{eq}\\ {/eq}

The standard form of quadratic equation and the quadratic formula are very useful in determining the solutions or roots of a quadratic equation. We will start with the coefficient calculation using the method of comparison then we will substitute these values into the formula and just simplify it and get the solutions. When finding the solutions, we have to determine such values of {eq}x {/eq} for which the value of the function or {eq}y {/eq} becomes equal to zero.

$$ax^{2} + bx + c = 0 \\[0.2cm] x = \Biggr[ \dfrac {-b \pm \sqrt {b^{2} - 4ac}}{2a} \Biggr] $$

Answer and Explanation:

{eq}\\ {/eq}

The polynomial for which we have to determine the solutions is given below:

{eq}y = 2x^{2} - 4x + 1 \\ {/eq}

Setting {eq}y = 0 {/eq}:

$$\begin{align} 2x^{2} - 4x + 1 &= 0 &\left( \text{ 1 }\right) \end{align} $$

Now compare the above equation with the standard one:

$$\begin{align} &ax^{2} + bx + c = 0 \\[0.2cm] &a = 2, \; b = -4, \; c = 1 \\ \end{align} $$

Substituting the values,

$$\begin{align} x &= \dfrac {-b \pm \sqrt {b^{2} - 4ac}}{2a} \\[0.2cm] x &= \dfrac {-(-4) \pm \sqrt {(-4)^{2} - 4 \times 2 \times 1}}{2 \times 2} \\[0.2cm] x &= \dfrac {4 \pm \sqrt {16 - 8}}{4} \\[0.2cm] x &= \dfrac {4 \pm \sqrt {8}}{4} \\[0.2cm] x &= \dfrac {4 \pm 2 \sqrt {2}}{4} \\[0.2cm] x &= 1 \pm \biggr( \dfrac {1}{\sqrt {2}} \biggr) \\[0.2cm] \end{align} \\ $$

Therefore, the roots are {eq}x = 1 + \biggr( \dfrac {1}{\sqrt {2}} \biggr) {/eq} and {eq}x = 1 - \biggr( \dfrac {1}{\sqrt {2}} \biggr) {/eq}.


Learn more about this topic:

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How to Use the Quadratic Formula to Solve a Quadratic Equation

from Math 101: College Algebra

Chapter 4 / Lesson 10
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