find roots of quadratic equation 3√2x²-5x-√2=0
Answers
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
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Answer:
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.