Twice the difference of a number and 2 is equal to three times the sum of the number and 4

Mathematics

1 answer:

ikadub [295]3 years ago

5 0

Answer:

Step-by-step explanation:

We are given that Twice the difference of a number and 2 is equal to three times the sum of the number and 4. And we have to tell the number.

Let the number be x

Its difference from 2 = x-2

Twice the difference =2(x-2)

Sum of number and 4 = x+4

Three times of the sum : 3(x+4)

Hence 2(x+2)=3(x+4)

2x+2×2=3x+3×4

2x+4=3x+12

Subtracting 12 and 2x from both sides.

4-12=3x-2x

-8=x

Hence the number is -8

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To find the zeros of a quadratic fiunction given the equation you can use the next quadratic formula after equal the function to 0:

$\begin{gathered} ax^2+bx+c=0 \\ \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \end{gathered}$

For the given function:

$f(x)=2x^2-10x-3$ $x=\frac{-(-10)\pm\sqrt[]{(-10)^2-4(2)(-3)}}{2(2)}$ $x=\frac{10\pm\sqrt[]{100+24}}{4}$ $\begin{gathered} x=\frac{10\pm\sqrt[]{124}}{4} \\ \\ x=\frac{10\pm\sqrt[]{2\cdot2\cdot31}}{4} \\ \\ x=\frac{10\pm\sqrt[]{2^2\cdot31}}{4} \\ \\ x=\frac{10\pm2\sqrt[]{31}}{4} \\ \end{gathered}$ $\begin{gathered} x_1=\frac{10}{4}+\frac{2\sqrt[]{31}}{4} \\ \\ x_1=\frac{5}{2}+\frac{\sqrt[]{31}}{2} \end{gathered}$ $\begin{gathered} x_2=\frac{10}{4}-\frac{2\sqrt[]{31}}{4} \\ \\ x_2=\frac{5}{2}-\frac{\sqrt[]{31}}{2} \end{gathered}$

Then, the zeros of the given quadratic function are: