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Mathematics

1 answer:

vovikov84 [41]3 years ago

4 0

Answer:

$\boxed{2\pm \frac{\sqrt{2}}{2}}$

Step-by-step explanation:

$2x^2-8x=-7$

$\sf Add \ 7 \ on \ both \ sides}.$

$2x^2-8x+7=-7+7$

$2x^2-8x+7=0$

$ax^2 +bx+c=0$

$\sf Apply \ quadratic \ formula.$

$a=2 \ \ \ b=-8 \ \ \ c = 7$

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

$x=\frac{-(-8)\pm\sqrt{(-8)^2-4(2)(7)}}{2(2)}$

$x=\frac{8\pm\sqrt{64-56}}{4}$

$x=\frac{8\pm\sqrt{8}}{4}$

$x=\frac{8\pm2\sqrt{2}}{4}$

$x=2\pm \frac{\sqrt{2}}{2}$

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2. Your friend wants to write the equation of a quadratic function 1 5. 3.r = 1) that has zeros at 3 and 3. They want to use the

Feliz [49]

Let's expand both options, your friend's and yours, as shown below

$\begin{gathered} y=(3x+1)(3x-5)=9x^2-12x-5=9(x^2-\frac{4}{3}x-\frac{5}{9}) \\ \text{and} \\ y=(x+\frac{1}{3})(x-\frac{5}{3})=x^2-\frac{4}{3}-\frac{5}{9} \end{gathered}$

Then, both equations are the same besides a constant that will not affect the zeros of the functions, as shown below

$\begin{gathered} y=(3x+1)(3x-5) \\ x=-\frac{1}{3} \\ \Rightarrow y=(-1+1)(-1-5)=0 \\ \text{and} \\ x=\frac{5}{3} \\ \Rightarrow y=(5+1)(5-5)=0 \\ \end{gathered}$

And

$\begin{gathered} y=(x-\frac{5}{3})(x+\frac{1}{3}) \\ x=-\frac{1}{3} \\ \Rightarrow y=(-\frac{1}{3}-\frac{5}{3})\cdot0=0 \\ x=\frac{5}{3} \\ \Rightarrow y=0\cdot(\frac{6}{3})=0 \end{gathered}$

Both your friend and you are correct. The functions are the same with exception of a constant that multiplies the whole function (a scale factor); despite that, the zeros are the same for both functions