$Suppose~that~the~number~is~x.\\According~to~question,\\x^2-140=3x\\or, x^2-3x-140=0\\From~the~quadratic~formula,\\x = \frac{-(-3)+\sqrt{(-3)^2-4(1)(-140)} }{2(1)} ~and ~x= \frac{-(-3)-\sqrt{(-3)^2-4(1)(-140)} }{2(1)} \\or, x = \frac{3+\sqrt{9+560} }{2}~and~x=\frac{3-\sqrt{9+569} }{2}\\or, x = \frac{3+\sqrt{569} }{2}~and~x=\frac{3-\sqrt{569} }{2}\\So,~the~required~negative~number~is~\frac{3-\sqrt{569} }{2}.$

5 0

2 years ago

What is the perimeter of the quadrilateral shown below?

GarryVolchara [31]

The perimeter of a shape is the sum of the lengths of its sides.

So, to find the perimeter of this quadrilateral, all we have to do is add the side lengths and simplify.

(x² - 6) + (2x + 5) + (x² - 3x) + (4x² + 2x)

x² + x² + 4x² + (-3x) + 2x + 2x + (-6) + 5

6x² + (-3x) + 2x + 2x + (-6) + 5

6x² + x + (-6) + 5

6x² + x + (-1)

6x² + x - 1

So, the perimeter of the quadrilateral is the quantity (6x² + x - 1).

Hope this helps!

8 0

2 years ago