Question

solve the equation 2x^3 +3x^2-8x+3=0 given that one root is double the other root​

Answer 1

Answer:

x = √-1

Step-by-step explanation:

→2x³ + 3x² - 8x + 3 = 0

→2x³ - 5x = 0 - 3

→ -3x² = -3

→ x² = -3/-3

→ x² = -1

→ x = √-1

Ans. x = √-1

Answer 2

Given,

Equation is $2x^{3}+3x^{2} -8x+3=0$

One root is double the other

To find,

Solving the equation

Solution,

$2x^{3}+3x^{2} -8x+3=0$  

Since it is an equation with power 3, it will have 3 roots.

Let one of the root be z.

As the other root is double of the root , it will be 2z.

Let the third root be y.

General cubic equation is , $ax^{3}+bx^{2} -cx+d=0$

By comparing given equation with general cubic equation, we have

a = 2

b = 3

c = -8

d = 3

We know that, sum of the roots is equal to $-\frac{b}{a}$

So, z+2z+y = $-\frac{3}{2}$

⇒y = $-\frac{3}{2}$ - 3z (Equation 1)

We know that, product of 2 roots at a time = $\frac{c}{a}$

So, $2z^{2}+2zy+zy = \frac{-8}{2}$

⇒2$z^{2}$ + 3zy = -4 (Equation 2)

Product of all roots = $-\frac{d}{a}$

So, $2z^{2}y$ = $-\frac{3}{2}$  (Equation 3)

So, take equation 2 ,  2$z^{2}$+3zy=-4

Substitute 1 in 2,

2$z^{2}$+3z($-\frac{3}{2}$$-3z$) = -4

$4z^{2} - 9z -18z^{2} = -8$

$14z^{2}+9z-8=0$

Factorize,

$14z^{2} + 16z-7z-8=0$

$(2z-1)(7z+8)=0\\z=\frac{1}{2}\\z=\frac{-8}{7}$

Now substitute z = $\frac{1}{2}$ in equation 1

y = $-\frac{3}{2}-\frac{3}{2} = -\frac{6}{2} = -3$

Now substitute z = $-\frac{8}{7}$

y = $-\frac{3}{2} +-\frac{24}{7} = -\frac{27}{14}$

Now lets check which z value to consider,

substitute z = $\frac{1}{2}$ in equation 3,

LHS = RHS hence we should consider this z.

So when we substitute z =$\frac{-8}{7}$ ,

LHS ≠ RHS

Hence the roots will be  $\frac{1}{2},1, -3$ .