In: Civil Engineering
Note:-
To solve this first you have to make a square of X2 and then place the square beside and below of X2 by dividing the equal value of the coefficient of X means here the coefficient is 6 hence (3+3), that's why I have drawn 3 squares besides of X2 and 3 squares below X2. Then u have to make another small square of last no, (here it is 8). then u have to fill a full square. which is indicated by a hatched line. Then u have to see how many squares are required to complete the whole square(here required square is =1). Hence the total length of 1 side of the large square is =X-3.
So, the total area of large square =(X-3)x(X-3)
but 1 square is missing here. then u have to subtract this small square from the large square.
Hence (X-3)x(X-3)-1=0
X=4 and X=2
14 Week 21 Solving aljebra th + 6x it8 by x + 6 x 48 tiles. X-3 fall square Need 1 squase to complete (8-3):1 = 0 =) - (x-3) = 1 (x-3) = 21 X-3=1 X = (1+3) X-3 = -1 x=(1+3) = X=2
14th Week 091- 275 Solving x + 6xit 8 by aljebra tiles. X +6x48 X - 3 XZ