HELP ASAP IF YOU TAKE THE POINTS IM REPORTING U QUESTION IN PIC

Mathematics

2 answers:

Andrew [12]3 years ago

7 0

Let's call the number x. So four times the number minus five would look like
4x-5
Twice the number plus three would look like
2x+3
Now make the two equations equal
4x-5=2x+3
I'm going to subtract 2x from each side
2x-5=3
Now I'm going to add five to each side
2x=8
Now divide both sides by 2
x=4
To double check, just plug back into the original equations
4(4)-5=11
2(4)+3=11
They are equal, so the answer is x=4.

Airida [17]3 years ago

4 0

Step-by-step explanation:

Let the number be x,

Then, according to the question,

4x-5=2x+3

(we need to find the value of x)

4x-2x=5+3

= 2x=8

= x=8/2

= x=4

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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: