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creativ13 [48]
3 years ago
11

Find the number of real number solutions for the equation. x2 + 5x + 7 = 0 0 cannot be determined 1 2

Mathematics
2 answers:
alisha [4.7K]3 years ago
5 0

\Delta=5^2-4\cdot1\cdot7=25-28=-3

\Delta so 0 solutions.

DaniilM [7]3 years ago
4 0

Answer:

No Real roots to this Quadratic Equation

Step-by-step explanation:

Our Quadratic equation is given as

x^2+5x+7=0

In order to find that do we have the real roots of a quadratic equation , the Discriminant must be greater or equal to 0. The Discriminant is denoted by D and given by the formula

D= b^2-4ac

Where b is the coefficient of the middle term containing x, a is the coefficient of the term containing x^{2} and the c is the constant term.

Hence we have

a = 1 , b = 5 and c = 7

Calculate D

D=b^2-4ac\\D=5^2-4*1*7\\D=25-28\\D=-3

Hence we see that the Discriminant (D) is less than 0, our answer is no real roots to this quadratic equation.

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A sailboat and a power boat leave from the same marina at the same time. The sailboat travels at 7 km/h, 14? west of south. The
GREYUIT [131]
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side b = 6 times 7
side c = 6 times 19
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Determime the intercepts of the line that passes through the following points. (-6,5);(-3,3);(0,1)
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\bf (\stackrel{x_1}{-6}~,~\stackrel{y_1}{5})\qquad
(\stackrel{x_2}{0}~,~\stackrel{y_2}{1})
\\\\\\
slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{1-5}{0-(-6)}\implies \cfrac{1-5}{0+6}\implies -\cfrac{5}{6}


\bf \begin{array}{|c|ll}
\cline{1-1}
\textit{point-slope form}\\
\cline{1-1}
\\
y-y_1=m(x-x_1)
\\\\
\cline{1-1}
\end{array}\implies y-5=-\cfrac{5}{6}[x-(-6)]\implies y-5=-\cfrac{5}{6}(x+6)
\\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
\textit{to get the x-intercept, we set y = 0, solve for \underline{x}}
\\\\\\
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\\\\\\
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Stels [109]

Factorize the quadratic trinomial x^2 - 9x + 14 by the rule:

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2. Factorize the polynomial:

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3. Only factor x-2 is given in options, then the correct choice is B.

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