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Anna007 [38]
3 years ago
10

How to solve 12ax-2ax+14x-2a+ -3x

Mathematics
1 answer:
Advocard [28]3 years ago
8 0
That long, messy expression is not an equation and it's not a question,
so there's nothing to solve or answer.  However, it IS long and messy, so
your homework assignment might be to 'simplify' it.  Here's how:

<span>12ax - 2ax + 14x - 2a + -3x

Everything we'll do to this is called "combine like terms".  That means
"combine terms that are alike".  Terms are alike when they have the
same letter variables.
 
12ax  and  -2ax  are like terms.  Combine them, and you get  10ax .

14x  and  -3x  are like terms.  Combine them, and you get  11x .

So now we have:  10ax + 11x + 2a .

Nothing has been changed.  This shorter, simpler expression has
exactly the same value as the original long messy one has.  It has
just been cleaned up, and it looks neater and simpler.

You might also be expected to 'factor' the expression.  I don't know.
If you want to do that, then you could do it in 2 different ways . . .
either factor the 'x' from the two terms that have 'x' in them, or
else factor the 'a' from the two terms that have 'a' in them. 
Without explanation, here's how those would look:

           x(10a + 11) + 2a
or     
           a(10x + 2) + 11x .

Again, nothing has changed.  Both of these have the same value, and
their value is the same as the value of the long messy expression in
the question.
 
</span>
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The quadratic equation 8x²+12x-14 has two real roots. What is the sum of the squares of these roots?
mars1129 [50]

Answer:

The real roots are

x=\frac{(-3+\sqrt{37})}{4} and x=\frac{(-3-\sqrt{37})}{4}

The sum of the squares of these roots is \frac{-3}{2}

Step-by-step explanation:

The given quadratic equation is 8x^2+12x-14 has two real roots.

To find the roots .

8x^2+12x-14=0

Dividing the above equation by 2

\frac{1}{2}(8x^2+12x-14)=\frac{0}{2}

4x^2+6x-7=0

For quadratic equation ax^2+bx+c=0 the solution is x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

Where a and b are coefficents of x^2 and x respectively, c is a constant.

For given quadratic equation

a=4, b=6, c=-7

x=\frac{-6\pm\sqrt{6^2-4(4)(-7)}}{2(4)}

=\frac{-6\pm\sqrt{36+112}}{8}

=\frac{-6\pm\sqrt{148}}{8}

=\frac{-6\pm\sqrt{37\times 4}}{8}

=\frac{-6\pm\sqrt{37}\times\sqrt{4}}{8}

=\frac{-6\pm\sqrt{37}\times 2}{8}

=2\frac{(-3\pm\sqrt{37})}{8}

=\frac{-3\pm\sqrt{37}}{4}

x=\frac{(-3\pm\sqrt{37})}{4}

The real roots are

x=\frac{(-3+\sqrt{37})}{4} and x=\frac{(-3-\sqrt{37})}{4}

Now to find the sum of the squares of these roots

\left[\frac{-3+\sqrt{37}}{4}+\frac{(-3-\sqrt{37})}{4}\right]^2=\frac{-3+\sqrt{37}-3-\sqrt{37}}{4}

=\frac{-6}{4}

=\frac{-3}{2}

\left[\frac{-3+\sqrt{37}}{4}+\frac{(-3-\sqrt{37})}{4}\right]^2=\frac{-3}{2}

Therefore the sum of the squares of these roots is \frac{-3}{2}

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Answer:

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Step-by-step explanation:

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