Question

11. Find the roots of the equation 2x^2– 5x + 3 = 0, by method of competing the square.

Class:- 10

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Answer 1

2x^2 - 5x +3 = 0

Step 1.) Transfer the constant term in RHS.

=> 2x^2 - 5x = - 3

Step 2.) Divide the whole equation by the coficient of x^2

=> x^2 - 5x/2 = - 3/2

Step 3.) Add the square of half of the coficient of x in both sides.

=> x^2 - 5x/2 + (5/4)^2 = - 3/2 + (5/4)^2

=> ( x - 5/4)^2 = - 3/2 + 25 / 16

=> (x - 5/4) ^2 = ( - 24 + 25) / 16

=> (x - 5/4)^2 = 1 / 16

=> x - 5/4 = plus minus 1 / 4

=> x = 5 / 4 plus minus 1/4

Alpha = 5/4 + 1/4 = 6/4 = 3/2

Beta = 5/4 - 1/4 = 4 /4 = 1

Roots are 3 /2 and 1

Answer 2

Hey!

$2x {}^{2} - 5x + 3 = 0$

$2x {}^{2} - 5x = - 3$

divide both sides of equation by 2

$x {}^{2} - \frac{5}{2} x + (\frac{5}{4} ) {}^{2} = \frac{ - 3}{2} + (\frac{5}{4} ) {}^{2}$

using a² - 2ab + b²  = (a - b)² 

$(x - \frac{5}{4} ) {}^{2} = \frac{1}{16}$

$x - \frac{5}{4} = + - \frac{1}{4}$

$x - \frac{5}{4} = \frac{1}{4}$

$x = \frac{3}{2}$

$x - \frac{5}{4} = \frac{ - 1}{4}$

$x = 1$

So, the two roots of the equation are 3/2 and 1

Hope it helps!