Answer: Two consecutive square numbers be 16 and 9.
Step-by-step explanation:
Let first consecutive square number be x²
Let second consecutive square number be (x+1)²
According to question,
[tex]x^2\times (x+1)^2=144[/tex]
Now we'll do this by factorisation,
[tex]144=2\times2\times2\times2\times3\times3\\\\144=16\times 9[/tex]
So, two consecutive square numbers are 16 and 9
Consecutive numbers are numbers that follow one another.
The consecutive squares are: 9 and 16
Let the bigger square be x^2
So, we have:
[tex]\mathbf{(x - 1)^2 \times x^2 = 144}[/tex]
Express 144 as 12^2
[tex]\mathbf{(x - 1)^2 \times x^2 = 12^2}[/tex]
Take square roots of both sides
[tex]\mathbf{(x - 1)\times x = 12}[/tex]
Expand
[tex]\mathbf{x^2 - x = 12}[/tex]
Rewrite as:
[tex]\mathbf{x^2 - x - 12 = 0}[/tex]
Expand
[tex]\mathbf{x^2 - 4x +3x- 12 = 0}[/tex]
Factorize
[tex]\mathbf{x(x - 4) +3(x- 4) = 0}[/tex]
Factor out x - 4
[tex]\mathbf{(x +3) (x- 4) = 0}[/tex]
Solve for x
x= -3 or x = 4
x cannot be negative.
So: x = 4
Square both sides
[tex]\mathbf{x^2 = 4^2}[/tex]
[tex]\mathbf{x^2 = 16}[/tex]
Calculate the smaller square
[tex]\mathbf{(x - 1)^2 = (4-1)^2}[/tex]
[tex]\mathbf{(x - 1)^2 = 9}[/tex]
Hence, the consecutive squares are: 9 and 16
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