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Stolb23 [73]
2 years ago
7

What is the value of "c" in the following quadratic? (Make sure the equation is in

Mathematics
1 answer:
garri49 [273]2 years ago
5 0

               \rule{50}{1}\large\blue\textsf{\textbf{\underline{Given question:-}}}\rule{50}{1}

           <em>What is the value of c in the quadratic </em>\large\text{$x^2+28=-11x$}?

          \rule{50}{1}\large\textsf{\textbf{\underline{Answer and how to solve:-}}}\rule{50}{1}

           Before starting to solve, you should notice something - the

quadratic is not in its standard form!

We can easily fix it by adding \large\textit{11x} on both sides:-

\large\text{$x^2+28-11x=0$}

We can switch the order of 28 and -11x:-

\large\text{$x^2-11x+28=0$}

  Now, the quadratic is in its standard form, so we can get down to

finding the value of "c".

Remember, the standard form of a quadratic looks like so:-

  •  \large\text{$ax^2+bx+c=0$}

Now we can just write our <u>quadratic</u> here:-

  • \large\text{$x^2-11x+28=0$}

Now, can you see what the value of "c" is?

An easy way to <u>remember</u> "c" in quadratics is:-

The "c" in quadratics is the constant.

   

Henceforth, we conclude that the value of "c" in the given quadratic is:-

 \Large\textbf{28}\Large\checkmark

<h3>         Good luck with your studies.</h3>

        \rule{50}{1}\smile\smile\smile\smile\smile\smile\rule{50}{1}    

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Answer:

<h3><u>Let's</u><u> </u><u>understand the concept</u><u>:</u><u>-</u></h3>

Here angle B is 90°

So \triangle ABC and \triangle ABD Are right angled triangle

So we use Pythagoras thereon for solution

<h3><u>Required Answer</u><u>:</u><u>-</u></h3>
  • First in triangle ABC

perpendicular=p=8cm

Hypontenuse =h =10cm

  • We need to find base=b

According to Pythagoras thereon

{\boxed{\sf b^2=h^2-p^2}}

  • Substitutethe values

\longrightarrow\sf b^2=10^2-p^2

\longrightarrow\sf b={\sqrt {10^2-8^2}}

\longrightarrow\sf b={\sqrt{100-64}}

\longrightarrow\bf b={\sqrt {36}}

\longrightarrow\sf b=6

\therefore\overline{BC}=6cm

  • BD=BC+CD

\longrightarrowBD=9+6

\longrightarrowBD=15cm

  • Now in \triangle ABD

Perpendicular=p=8cm

Base =b=15cm

  • We need to find Hypontenuse =AD(x)

According to Pythagoras thereon

{\boxed {\sf h^2=p^2+b^2}}

  • Substitute the values

\longrightarrow\sf h^2=8^2+15^2

\longrightarrow\sf h={\sqrt {8^2+15^2}}

\longrightarrow\sf h={\sqrt {64+225}}

\longrightarrow\sf h={\sqrt {289}}

\longrightarrow\sf h=17cm

\therefore{\underline{\boxed{\bf x=17cm}}}

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