Question

How do you solve \[4{x^2} - 13x + 3 = 0\]?

Answer 1

Hint: From the given question, we can see that the equation is a quadratic equation. When an equation is in the form of \[a{x^2} + bx + c = 0\], it is a quadratic equation. Here we will use the quadratic formula that is \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] to solve our question.

Complete step by step solution:
According to the question, the given equation is:
\[4{x^2} - 13x + 3 = 0\]
We know that the above equation is in the form of a quadratic equation. The form of the quadratic equation is:
\[a{x^2} + bx + c = 0\]
When we relate it with our given equation, we get:
\[a = 4\,;\,b = - 13\,;\,c = 3\]
Now, we will use the quadratic formula to solve the equation, the formula is:
\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] where a, b and c are constants.
By putting the values of a, b and c in the equation, we get:
 \[ \Rightarrow x = \dfrac{{13 \pm \sqrt {{{( - 13)}^2} - 4 \times 4 \times 3} }}{{2 \times 4}}\]
Now, we will try to solve the equation here, and we will get:
\[ \Rightarrow x = \dfrac{{13 \pm \sqrt {169 - 48} }}{8}\]
\[ \Rightarrow x = \dfrac{{13 \pm \sqrt {121} }}{8}\]
After taking out the square root, we get:
 \[ \Rightarrow x = \dfrac{{13 \pm 11}}{8}\]
Now, we will solve for ‘x’. We will get two values of ‘x’ because it is a quadratic equation:
\[x = \dfrac{{13 + 11}}{8}\] and \[x = \dfrac{{13 - 11}}{8}\]
When we solve for \[x = \dfrac{{13 + 11}}{8}\], we get:
\[ \Rightarrow x = \dfrac{{24}}{8}\]
After cancelling or dividing the numerator and denominator, we get:
\[ \Rightarrow x = 3\]
When we solve for \[x = \dfrac{{13 - 11}}{8}\], we get:
\[ \Rightarrow x = \dfrac{2}{8}\]
 After cancelling or dividing the numerator and denominator, we get:
\[ \Rightarrow x = \dfrac{1}{4}\]or \[x = 0.25\]
Therefore, after solving, we get the values of ‘x’ as \[x = 3\,\,\dfrac{1}{4}\].

Note: The above method is very easy, and we can solve the question very quickly. But there is another method to solve this question. We can use the new Transforming Method here. This method is much simpler and faster than the above method where we use the formulas.