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3 years ago

9

Determine whether triangle TJD is congruent to triangle SEK given T (-4,-2), J (0,5), D (1,-1), S (-1,3), E (3,10), K (4,4) and

explain?

1 answer:

nataly862011 [7]3 years ago

7 0

Yes, it is.

The three points of triangle SEK are the points of TJD shifted 3 units right and 5 units upward.

So they are the same triangle, just translated in the plane.

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2x2 – 5x + 67 = 0<br> What would be your first step in completing the square for the equation above?

7nadin3 [17]

Answer:

The first step is to divide all the terms by the coefficient of $x^{2}$ which is 2.

The solutions to the quadratic equation $2x^2\:-\:5x\:+\:67\:=\:0$ are:

$x=\frac{5}{4}+i\frac{\sqrt{511}}{4},\:x=\frac{5}{4}-i\frac{\sqrt{511}}{4}$

Step-by-step explanation:

Considering the equation

$2x^2\:-\:5x\:+\:67\:=\:0$

The first step is to divide all the terms by the coefficient of $x^{2}$ which is 2.

so

$\frac{2x^2-5x}{2}=\frac{-67}{2}$

$x^2-\frac{5x}{2}=-\frac{67}{2}$

Lets now solve the equation by completeing the remaining steps

Write equation in the form: $x^2+2ax+a^2=\left(x+a\right)^2$

Solving for $a$ ,

$2ax=-\frac{5}{2}x$

$a=-\frac{5}{4}$

$\mathrm{Add\:}a^2=\left(-\frac{5}{4}\right)^2\mathrm{\:to\:both\:sides}$

$x^2-\frac{5x}{2}+\left(-\frac{5}{4}\right)^2=-\frac{67}{2}+\left(-\frac{5}{4}\right)^2$

$x^2-\frac{5x}{2}+\left(-\frac{5}{4}\right)^2=-\frac{511}{16}$

Completing the square

$\left(x-\frac{5}{4}\right)^2=-\frac{511}{16}$

Since, you had required to know the first step in completing the square for the equation above, I hope you have got the point, but let me quickly solve the remaining solution.

For $f^2\left(x\right)=a$ the solution are $f\left(x\right)=\sqrt{a},\:-\sqrt{a}$

Solving

$x-\frac{5}{4}=\sqrt{-\frac{511}{16}}$

$x-\frac{5}{4}=\sqrt{-1}\sqrt{\frac{511}{16}}$

$x-\frac{5}{4}=i\sqrt{\frac{511}{16}}$ ∵ Applying imaginary number rule $\sqrt{-1}=i$

$x-\frac{5}{4}=i\frac{\sqrt{511}}{\sqrt{16}}$

$-\frac{5}{4}=i\frac{\sqrt{511}}{4}$

$x=\frac{5}{4}+i\frac{\sqrt{511}}{4}$

Similarly, solving

$x-\frac{5}{4}=-\sqrt{-\frac{511}{16}}$

$x-\frac{5}{4}=-i\frac{\sqrt{511}}{4}$ ∵ Applying imaginary number rule $\sqrt{-1}=i$

$x=\frac{5}{4}-i\frac{\sqrt{511}}{4}$

Therefore, the solutions to the quadratic equation are:

$x=\frac{5}{4}+i\frac{\sqrt{511}}{4},\:x=\frac{5}{4}-i\frac{\sqrt{511}}{4}$

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