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Hello can you tell me the last part pleas le

Hello can you tell me the last part pleas le-example-1
User Rune G
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We have the following coefficients of the quadratic function:

a = -10

b = 23

c = -12

And we need to find the roots to that function, using the formulas:


\begin{gathered} \frac{-b-\sqrt[]{b^(2)-4ac}}{2a} \\ \\ \frac{-b+\sqrt[]{b²-4ac}}{2a} \end{gathered}

In order to do so, we need to replace each constant with its numerical value. So, we obtain:


\begin{gathered} \frac{-b-\sqrt[]{b²-4ac}}{2a}=\frac{-23-\sqrt[]{23^(2)-4(-10)(-12)}}{2(-10)} \\ \\ =\frac{-23-\sqrt[]{529-4(120)}}{-20}\text{ since the product of two negative numbers is positive} \\ \\ =\frac{-23-\sqrt[]{529-480}}{-20} \\ \\ =\frac{-23-\sqrt[]{49}}{-20} \\ \\ =(-23-7)/(-20) \\ \\ =(-30)/(-20) \\ \\ =(3)/(2) \end{gathered}

Now, the second formula is similar to the first one: we only need to change the sign before the square root. Thus, we obtain:


\frac{-b+\sqrt[]{b²-4ac}}{2a}=(-23+7)/(-20)=(-16)/(-20)=(4)/(5)

User Derstauner
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