Question

Solve the equation $4{{x}^{2}}-5x-3=0$.

Answer 1

Hint: In this question, we are given a quadratic equation and we have to solve it which means we have to find the roots of the equation. Since the equation is of degree 2, so we will obtain two roots of this equation. We will use a quadratic formula to calculate the roots of this equation. For an equation $a{{x}^{2}}+bx+c$ the roots using quadratic formula are given by $\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ where $\sqrt{{{b}^{2}}-4ac}$ is known as discriminant.

Complete step by step answer:
Here, we are given an equation as $4{{x}^{2}}-5x-3=0$. We have to find its root. Given equation is a quadratic equation. Since, equation is of degree 2, therefore we will obtain two roots of this equation. Let us use a quadratic formula for calculation of roots. As we know, for any equation $a{{x}^{2}}+bx+c$ roots are given by $\Rightarrow x=\dfrac{-b\pm \sqrt{D}}{2a}$ where D is discriminant, equal to $D={{b}^{2}}-4ac$.
D determines the nature of roots.
If D>0 then we have real and distinct roots.
If D<0 then we have imaginary roots and
If D = 0 then we have equal real roots.
So, let us compare given equation $4{{x}^{2}}-5x-3=0$ by $a{{x}^{2}}+bx+c$ to obtain value of a, b and c. We get, a = 4, b = -5 and c = -3.
Let us first calculate D to determine the nature of roots. We get,
\[\begin{align}
  & D={{b}^{2}}-4ac \\
 & \Rightarrow {{\left( -5 \right)}^{2}}-4\left( 4 \right)\left( -3 \right) \\
 & \Rightarrow 25+48 \\
 & \Rightarrow 73 \\
\end{align}\]
Now, let us use quadratic formula to find value of x, $\Rightarrow x=\dfrac{-b\pm \sqrt{D}}{2a}$ so we get:
\[\begin{align}
  & \Rightarrow x=\dfrac{-\left( -5 \right)\pm \sqrt{73}}{2\left( 4 \right)} \\
 & \Rightarrow x=\dfrac{5\pm \sqrt{73}}{8} \\
 & \Rightarrow x=\dfrac{5+\sqrt{73}}{8}\text{ and }x=\dfrac{5-\sqrt{73}}{8} \\
\end{align}\]

Hence two roots of given equation are $\dfrac{5+\sqrt{73}}{8}\text{ and }\dfrac{5-\sqrt{73}}{8}$.

Note: Students should note that here, we have used a quadratic method because roots are irrational numbers. If roots were rational numbers, we could split the middle term method for solving this equation. While comparing the equation with $a{{x}^{2}}+bx+c$ take care of signs as well. While applying quadratic formula, students can make mistakes in positive and negative signs. The degree of the equation gives us an idea about the number of roots an equation has.