Name: Solving Free Fall Problem Using Quadratic Equation
Uploaded: 2019-07-19
Duration: 7 min 7 s
Description: In this video, we will learn how we can solve the problem of free fall effectively using the quadratic equation. Let's watch the video and learn.

Question

Find the roots of the quadratic equation: ${a^{2} x}^{2} - 3 a b x + {2 b}^{2} = 0$ (if they exist) by the method of completing the square.

Accepted Answer

Given equation: $a^{2} x^{2} - 3 a b x + 2 b^{2} = 0$

$\Rightarrow a^{2} x^{2} - 3 a b x = - 2 b^{2}$ [Transposing constant term to RHS]

Adding ${(\frac{3 b}{2})}^{2}$ both sides, we get

$\Rightarrow {(a x)}^{2} - 2 (a x) (\frac{3 b}{2}) + {(\frac{3 b}{2})}^{2} = - 2 b^{2} + {(\frac{3 b}{2})}^{2}$

In above equation, LHS is of the form $a^{2} - 2 a b + b^{2}$ which is the identity of $(a - b)^{2}$

$\Rightarrow {(a x - \frac{3 b}{2})}^{2} = - 2 b^{2} + {\frac{9 b}{4}}^{2}$

Subtracting the term in RHS using LCM, we get

${(a x - \frac{3 b}{2})}^{2} = \frac{- 8 b^{2} + 9 b^{2}}{4} = \frac{b^{2}}{4} = {(\frac{b}{2})}^{2}$

$\Rightarrow a x - \frac{3 b}{2} = \pm \frac{b}{2}$

$\Rightarrow a x = \frac{3 b}{2} \pm \frac{b}{2}$

$\Rightarrow x = \frac{3 b \pm b}{2 a}$

$\Rightarrow x = \frac{3 b + b}{2 a}$ or $x = \frac{3 b - b}{2 a}$

$\Rightarrow x = \frac{4 b}{2 a} = \frac{2 b}{a}$ or $x = \frac{2 b}{2 a} = \frac{b}{a}$

Hence, $x = \frac{2 b}{a}$ and $x = \frac{b}{a}$ are the roots of the given equation.

Find the roots of the quadratic equation: ${a^{2} x}^{2} - 3 a b x + {2 b}^{2} = 0$ (if they exist) by the method of completing the square.

Important Questions on Quadratic Equations

Learn and Prepare for any exam you want !

Find the roots of the quadratic equation: a2x2-3abx+2b2=0 (if they exist) by the method of completing the square.

Important Questions on Quadratic Equations

Learn and Prepare for any exam you want !

Find the roots of the quadratic equation: ${a^{2} x}^{2} - 3 a b x + {2 b}^{2} = 0$ (if they exist) by the method of completing the square.