Question

what are the steps for solving an equation with competing square method​

Answer 1

Answer:

Step-by-step explanation:

first we have to check whether the ax square term is a perfect square or not if the first term is not then we have to multiply the term with the same value of a then we have to to find the value of b and add the square and subtract the square.

by this we get a equation with the identity (a+)^2 which we have to take and simplify the other b^2 we took before to take the value for eg

5/2 whole square then make it 25l4 and subtract with the c term.

the value you get should be sended to the other side and then we have square root both the sides.

if it is 3x-5/2= root 4 then simplify the root and we have to send -5/2 to the other side and put a +or- sign for root 4

with this we get 2 values

3x=5/2 + 2=7l2

3x=5/2-2= 3/2

to get the value of x in the given values then we have to divide the 7/3 and 5/3 with 3.

mark it as the brainliest!!!!

Answer 2

Lets take an example for better understanding

3x^2 - 5x + 2 =0

=> x^2 - 5/3x + 2/3 = 0

( ÷ the whole equation by coefficient )

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

( now add square of the coefficient of x on both sides ie, here it is (1/2 × 5/3)^2 = 25/36 )

=> x^2 -5/3x + 25/36 = -2/3 + 25/36

=> (x - 5/6)^2 = -24/36 + 25/36

[ x^2 -5/3x + 25/36 = (x - 5/6)^2 using (a+b)^2 identity ]

=> (x - 5/6)^2 = 1/36

=> x -5/6 = +/-1/6 ( rooting both sides)

When +1/6

x= 1/6 + 5/6 = 6/6 = 1

When - 1/6

x= -1/6 + 5/6 = 4/6

Hope it helped....