Answer:
her solution isn't correct .
because she had done a slight mistake in her calculation.
i.e.
[tex]\sqrt{16}=4 \:\:not \:\:16[/tex]
Step-by-step explanation:
solution given:
x²-6x-7=0
making it a perfect square
we must make it on the form of (a+b)²or(a-b)²
so
x²-2*3*x+3²-3²-7=0
(x-3)²-9-7=0
(x-3)²-16=0
adding 16 on both aide
(x-3)²=16
doing square root on both side
[tex]\sqrt{(x-3)²}=\sqrt{16}[/tex]
x-3=±4
taking positive
x-3=4
adding 3 on both side
x=7
and.
taking negative
x-3=-4
adding 3 on both side
x=-4+3
x=-1
x=-1 or 7
