Processing Math: Done

From Förberedande kurs i matematik 1

The first step when we solve the second-degree equation is to complete the square on the left-hand side

y2+2y−15=(y+1)2−12−15=(y+1)2−16.

The equation can now be written as

(y+1)2=16

and has, after taking the square root, the solutions:

• y+1=16=4,   which gives y=−1+4=3,

• y+1=−16=−4,   which gives y=−1−4=−5.

A quick check shows that y=−5 and y=3 satisfy the equation:

• y = -5:  LHS=(−5)2+2(−5)−15=25−10−15=0=RHS,

• y = 3:  LHS=32+23−15=9+6−15=0=RHS.