A number exceeds to its square root is 6. the number is
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Hey,sup!
Let the number be x^2.
As per the question,
=>x^2-√x^2=6.
=>x^2 -x = 6.
=>x^2-x-6=0.
=>x^2-3x+2x-6=0.
=>x(x-3)+2(x-3)=0.
=>(x-3)(x+2)=0.
We'll discard x+2 as it gives -ve value
So,
=>x-3=0.
=>x=3
So the number is x^2 = 3^2 = 9.
Hope it helps.
Let the number be x^2.
As per the question,
=>x^2-√x^2=6.
=>x^2 -x = 6.
=>x^2-x-6=0.
=>x^2-3x+2x-6=0.
=>x(x-3)+2(x-3)=0.
=>(x-3)(x+2)=0.
We'll discard x+2 as it gives -ve value
So,
=>x-3=0.
=>x=3
So the number is x^2 = 3^2 = 9.
Hope it helps.
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