Question

find the values of m for which the equation 3x^2+mx +2=0 has equal roots.Also find the roots of the given equation

Answer 1

Answer

m= +- 2 root 6 and x=-0.813

Step-by-step explanation:

use b^2 - 4ac = 0. Since the roots are equal

and we get m as root 24 which is 2 root 6.

x= -b +- sqrt (b^2-4ac) / 2a

Answer 2

The value of m=±2√6; if quadratic equation has equal roots. Roots of two separate equations thus formed are √6/3 and -√6/3.

Given:

$\bf 3 {x}^{2} + mx + 2 = 0$

To find:

• The value/es of m for which the equation $\bf 3 {x}^{2} + mx + 2 = 0$ has equal roots.
• Also find the roots of the given equation.

Solution:

Concept to be used:

For equal roots D= 0.

$\bf D = {b}^{2} - 4ac$

if quadratic equation is written as

$\bf a {x}^{2} + bx + c = 0$

where,a≠0.

Step 1:

Write coefficients of x²,x and constant term.

$a = 3 \\ b = m \\ c = 2$

Step 2:

Calculate D=0

${m}^{2} - 4(3)(2) = 0$

or

${m}^{2} = 24$

or

$\bf m = \pm \: 2 \sqrt{6}$

Step 3:

As m have two values, there must be two quadratic equations.

Find the roots of quadratic equation.

As m= b

Case 1:

$\bf b = + 2 \sqrt{6}$

The quadratic equation thus formed is

$3 {x}^{2} + 2 \sqrt{6} x + 2 = 0$

Roots are equal,

Thus,

$x = \frac{ - 2 \sqrt{6} }{6}$

or

$\bf x_{1,2} = \frac{ - \sqrt{6} }{3}$

Case 2:

$b = 2 \sqrt{6}$

$3 {x}^{2} - 2 \sqrt{6} x + 2 = 0$

or

$\bf x_{1,2} = \frac{\sqrt{6} }{3}$

Thus,

Value of m=±2√6 and roots of two separate equations are +√6/3 and -√6/3.

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