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earnstyle [38]
2 years ago
12

Pls i need to pass this class i wna graduate

Mathematics
1 answer:
jasenka [17]2 years ago
7 0

Answer:

\boxed{\sf x= -29.5}

Step-by-step explanation:

\sf \sqrt{-2x+5}-3=5

\sf \sqrt{-2x+5}=8

\sf -2x+5=64

\sf -2x=59

\sf \cfrac{-2x}{-2}=\cfrac{59}{-2}

Therefore, x = -29.5.

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What number must you add to complete the square?<br> x^2 + 2x = -1
Andru [333]
\large\begin{array}{l} \textsf{You must add 1 to complete the square.}\\\\ \textsf{There is a very common special product: (square of a sum)}\\\\ \mathsf{(a+b)^2=a^2+2ab+b^2}\\\\\\ \textsf{Take the given equation:}\\\\ \mathsf{x^2+2x=-1\qquad(i)}\\\\\\ \textsf{The first term is a square (x squared)}\\\\ \textsf{The second term is a product of 2, x and 1.} \end{array}


\large\begin{array}{l}\\\\\\ \textsf{We need to find a proper value for b, so that if we add it to}\\\textsf{the left side of the equation, we get a perfect square.}\\\\ \textsf{So,}\\\\ \begin{array}{cccccccccc} \mathsf{a^2}&\!\!\!+\!\!\!&\mathsf{2}&\!\!\!\cdot\!\!\!&\mathsf{a}&\!\!\!\cdot\!\!\!&\mathsf{b}&\!\!\!+\!\!\!&\mathsf{b^2}&=\mathsf{(a+b)^2}\\\\ \mathsf{\downarrow}&&&\!\!\!\!\!&\downarrow&\!\!\!\!\!&\downarrow&&\downarrow\\\\ \mathsf{x^2}&\!\!\!+\!\!\!&\mathsf{2}&\!\!\!\cdot\!\!\!&\mathsf{x}&\!\!\!\cdot\!\!\!&\mathsf{1}&\!\!\!+\!\!\!&\mathsf{1^2}&=\mathsf{(x+1)^2} \end{array}\\\\\\ \textsf{Just by comparing the expressions above, we can conclude that}\\\textsf{b must be equal to 1, so we get a perfect square:}\\\\ \mathsf{(x+1)^2=x^2+2x+1} \end{array}


\large\begin{array}{l}\\\\\\ \textsf{Back to (i), adding 1 to both sides:}\\\\ \mathsf{x^2+2x+1=-1+1}\\\\ \boxed{\begin{array}{c} \mathsf{(x+1)^2=0} \end{array}} \end{array}


\large\begin{array}{l}\\\\\\ \textsf{If you want to solve this equation for x, you will find}\\\\ \mathsf{x+1=0}\\\\ \mathsf{x=-1}\quad\longleftarrow\quad\textsf{(solution)} \end{array}


If you're having problems understanding the answer, try seeing it through your browser: brainly.com/question/2112248


\large\begin{array}{l} \textsf{Any doubt? Please, comment below.}\\\\\\ \textsf{Best regards! :-)} \end{array}

6 0
3 years ago
Read 2 more answers
Find all the zeroes of the equation(with simple steps).
uysha [10]

<u>Answer-</u>

<em>The zeros are, 5,\ -5,\ 4i,\ -4i</em>

<u>Solution-</u>

\Rightarrow -3x^4+27x^2+1200=0

\Rightarrow -3(x^2)^2+27(x^2)+1200=0

Here,

a = -3, b = 27, c = 1200

So,

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

=\dfrac{-27\pm \sqrt{-27^2-4\cdot (-3)\cdot 1200}}{2\cdot (-3)}

=\dfrac{-27\pm \sqrt{729+14400}}{-6}

=\dfrac{-27\pm 123}{-6}

=\dfrac{-27+123}{-6},\ \dfrac{-27- 123}{-6}

=\dfrac{96}{-6},\ \dfrac{-150}{-6}

=-16,\ 25

So,

\Rightarrow x^2=25,\ -16

\Rightarrow x=\sqrt{25},\ \sqrt{-16}

\Rightarrow x=5,\ -5,\ 4i,\ -4i

8 0
3 years ago
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