Mn - 4m - 5n + 20

In this, mn and -4m have a common factor; m. So we can take that out to make it:

m(n-4) -5n + 20

Now, -5n + 20 also has a common factor which is 5, because 20/5 gives you 4, and 5/5 equals 1. So we can talk out that -5 (it doesn't need to be negative, but negatives on variable, no gusta). So you'd get

m(n-4)-5(n-4) OR m(n-4)+5(-n+4)

4 0

4 years ago

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Use the quadratic formula to solve the following equation -3x^2-x-3=0

tigry1 [53]

<u>Answer:</u>

$x=-\frac{1}{6}-\frac{\sqrt{35}}{6} i \text { and } x=-\frac{1}{6}+\frac{\sqrt{35}}{6} i$ are two roots of equation $-3 x^{2}-x-3=0$

<u>Solution:</u>

Need to solve given equation using quadratic formula.

$-3 x^{2}-x-3=0$

General form of quadratic equation is $a x^{2}+b x+c=0$

And quadratic formula for getting roots of quadratic equation is

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

In our case b = -1 , a = -3 and c = -3

Calculating roots of the equation we get

$\begin{array}{l}{x=\frac{-(-1) \pm \sqrt{(-1)^{2}-4(-3)(-3)}}{2 \times-3}} \\\\ {x=-\frac{1}{6} \pm\left(-\frac{\sqrt{-35}}{6}\right)}\end{array}$

Since $b^{2}-4 a c$ is equal to -35, which is less than zero, so given equation will not have real roots and have complex roots.

$\begin{array}{l}{\text { Hence } x=-\frac{1}{6}-\frac{\sqrt{35}}{6} i \text { and } x=-\frac{1}{6}+\frac{\sqrt{35}}{6} i \text { are two roots of equation - }} \\ {3 x^{2}-x-3=0}\end{array}$

8 0

4 years ago