Math, asked by ishaangarg2479, 8 months ago

Find roots of quadratic equation 3√2x²-5x-√2=0, wrong will be strictly reported

Answers

Answered by jayant2406
1

Answer:

Required zeroes of the given equation are √2 and - 1 / 3√2.

Step-by-step explanation:

Given quadratic equation : 3√2 x^2 - 5x - √2 = 0

Method 1

Using Quadratic Formula :

From the properties of quadratic equations, if an equation is ax^2 + bs + c = 0 then it's roots are :

+b±√b²-4ac/2a

On comparing the given equation with ax^2 + bx + c = 0, we get :

a = 3√2 ; b = - 5 ; c = - √2

Therefore,

= > x = [ - ( - 5 ) ± √{ ( - 5 )^2 - 4( 3√2 x - 2 ) } ] / ( 2 x 3√2 )

= > x = [ 5 ± √{ 25 + 24 } ] / ( 6√2 )

= > x = ( 5 ± √49 ) / 6√2

= > x = ( 5 ± 7 ) / 6√2

= > x = 12 / 6√2 Or - 2 / 6√2

= > x = √2 Or - 1 / 3√2

Method 2

Factorisation :

= > 3√2 x^2 - 5x - √2 = 0

Splitting the middle term( term having x ) so that the product of parts is equal to the product of coefficient of x^2 and the remaining term.

So here product of parts should be 3√2 x √2 = 6.

Required parts should be 6 and 1, since their product is 6.

= > 3√2 x^2 - ( 6 - 1 )x - √2 = 0

= > 3√2 x^2 - 6x + x - √2 = 0

= > 3√2x( x - √2 ) + ( x - √2 ) = 0

= > ( x - √2 )( 3√2 x + 1 ) = 0

= > x = √2 Or x = - 1 / 3√2

Hence the required zeroes of the given equation are √2 and - 1 / 3√2.

HOPE THIS HELPS YOU

KEEP STUDYING HARD

Answered by kush193874
4

Answer:

Step-by-step explanation:

3√2x²-5x-√2=0

splitting the middle term

3√2x²-(6-1)x-√2=0

3√2x²-6x+x-√2=0

3√2x(x-√2)+1(x-√2)=0

(x-√2)(3√2x-1)=0

case1. case2

x-√2=0. 3√2x-1=0

x=√2. x=1/3√2

if it was helpful plz mark me as brainliest

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