Question

How do you factor and solve $4{x^2} - 8x + 1 = 0$?

Answer 1

Hint: In order to determine the solution to the above quadratic equation having variable $x$, first we compare it with the standard equation i.e. $a{x^2} + bx + c$ to obtain the value of the coefficients. Then find the determinant to find out the nature of roots, in our case it comes to be $D > 0$ which means roots are distinct and real and now apply the quadratic formula $x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$ to get your required solutions.

Complete step by step answer:
Given a quadratic equation, ${x^2} - 8x + 15 = 0$ let it be $f(x)$and the roots to the equation be ${x_1},{x_2}$.
$f(x) = 4{x^2} - 8x + 1 = 0$
Comparing the equation with the standard Quadratic equation $a{x^2} + bx + c$
a becomes 4
b becomes -8
And c becomes 1
Let’s now find the determinant value of this quadratic equation, as by looking at the value of determinant we can find out the nature of both the roots.
$D = {b^2} - 4ac$
Putting the values, we get
$
  D = {\left( { - 8} \right)^2} - 4(1)(4) \\
   = 64 - 16 \\
   = 48 \\
 $
As we can see $D > 0$, so we can conclude that the both roots of the equation will be distinct and real.
Now using Quadratic formula to find out the actual value of the roots
x = $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Since, ${b^2} - 4ac = D$
$
  x = \dfrac{{ - b \pm \sqrt D }}{{2a}} \\
   \Rightarrow {x_1} = \dfrac{{ - b + \sqrt D }}{{2a}},{x_2} = \dfrac{{ - b - \sqrt D }}{{2a}} \\
    \\
 $ Now putting the value of D, b, a, we get
$
   \Rightarrow {x_1} = \dfrac{{ - ( - 8) + \sqrt {48} }}{{2(4)}},{x_2} = \dfrac{{ - ( - 8) - \sqrt {48} }}{{2(4)}} \\
   \Rightarrow {x_1} = \dfrac{{8 + \sqrt {48} }}{8},{x_2} = \dfrac{{8 - \sqrt {48} }}{8} \\
 $
Simplifying more further,$\sqrt {48} = 4\sqrt 3 $
$
   \Rightarrow {x_1} = \dfrac{{8 + 4\sqrt 3 }}{8},{x_2} = \dfrac{{8 - 4\sqrt 3 }}{8} \\
   \Rightarrow {x_1} = 4\left( {\dfrac{{2 + \sqrt 3 }}{8}} \right),{x_2} = 4\left( {\dfrac{{2 - \sqrt 3 }}{8}} \right) \\
   \Rightarrow {x_1} = \dfrac{{2 + \sqrt 3 }}{2},{x_2} = \dfrac{{2 - \sqrt 3 }}{2} \\
 $
Therefore, the solution to the quadratic equation $4{x^2} - 8x + 1 = 0$ having roots ${x_1},{x_2}$ is ${x_1} = \dfrac{{2 + \sqrt 3 }}{2},{x_2} = \dfrac{{2 - \sqrt 3 }}{2}$.

Additional Information:
Quadratic Equation: A quadratic equation is an equation which can be represented in the form of $a{x^2} + bx + c$ where $x$ is the unknown variable and a,b,c are the numbers known where $a \ne 0$. If $a = 0$ then the equation will become a linear equation and will no longer be quadratic.
The degree of the quadratic equation is of the order 2.
Every Quadratic equation has 2 roots.

Discriminant: $D = {b^2} - 4ac$
Using Discriminant, we can find out the nature of the roots
If D is equal to zero, then both of the roots will be the same and real.
If D is a positive number then, both of the roots are real solutions.
If D is a negative number, then the root are the pair of complex solutions

Note:
You can alternatively find out the solution to the given quadratic equation by using splitting up the middle term method.
Step 1: calculate the product of coefficient of ${x^2}$ and the constant
Step 2: Second Step is to find the 2 factors of the number 2 such that the whether addition or subtraction of those numbers is equal to the middle term or coefficient of x and the product of those factors results in the value of constant.
Step 3: Pick out common from the terms to make the expression in the form the form of factors.
Step 4: Put all the factors equal to zero to determine the value of variable $x$.