Question

how to solve the given Quadratic equation. 1960x² -39.2x - 15.68 = 0 ​

Answer 1

Answer:

Given the following quadratic equation.

=> 1960x² - 39.2x - 15.68 = 0

The HCF of each coefficient and constant is 392. So divide both sides by 392, we get.

=> 5x² - 0.1x - 0.04 = 0

To make calculation easier, we should not have the coefficient of x and the constant term in decimal. So Multiply both sides by 100.

=> 500x² - 10x - 4 = 0

Splitting the middle term, we get

=> 500x² - 50x + 40x - 4 = 0

=> 50x (10x - 1) + 4(10x - 1) = 0

=> (10x - 1)(50x + 4) = 0

Case-1

=> 10x - 1 = 0

=> 10x = 1

=> x = 1/10

Case-2

=> 50x + 4 = 0

=> 50x = -4

=> x = -4/50

=> x = -2/25

Hence, x = 1/10, -2/25

Answer 2

Topic :-

Quadratic Equation

To Solve :-

$\mathtt {1960x^2-39.2x-15.68=0}$

Solution :-

Technique 1

Using Quadratic Formula

The solution of the quadratic equation, ax² + bx + c = 0 is given by

$\mathtt{x=\dfrac{-b \pm \sqrt {b^2-4ac}}{2a}}$

Quadratic Equation : $\mathtt {1960x^2-39.2x-15.68=0}$

here

a = 1960

b = -39.2

c = -15.68

Apply Quadratic Formula,

$\mathtt{x=\dfrac{39.2 \pm \sqrt {(-39.2)^2-4(1960)(-15.68)}}{2(1960)}}$

$\mathtt{x=\dfrac{39.2 \pm \sqrt {(1536.64)+(122931.2)}}{3920}}$

$\mathtt{x=\dfrac{39.2 \pm \sqrt {124467.84}}{3920}}$

$\mathtt{x=\dfrac{39.2 \pm 352.8}{3920}}$

$\mathtt{x=\dfrac{39.2 + 352.8}{3920},\dfrac{39.2 - 352.8}{3920}}$

$\mathtt{x=\dfrac{392}{3920},\dfrac{-313.6}{3920}}$

$\mathtt{x=0.1,-0.08}$

Technique 2 :-

$\mathtt {1960x^2-39.2x-15.68=0}$

$\mathtt {1960x^2-196x+156.8x-15.68=0}$

$\mathtt {196x(10x-1)+15.68(10x-1)=0}$

$\mathtt {(10x-1)(196x+15.68)=0}$

$\mathtt {(10x-1)=0}$

$\mathtt {x=0.1}$

$\mathtt {(196x+15.68)=0}$

$\mathtt {x=-0.08}$

Answer :-

So, value of x which satisfies equation are 0.1 and -0.08.