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4 years ago

If x2 +8x-4=x2+8, then 2x-3+?

Mathematics

1 answer:

DanielleElmas [232]4 years ago

3 0

Answer: The number 0

x^2 + 8x - 4 = x^2 + 8

8x - 4 = 8 ... subtract x^2 from both sides

8x = 12 .... add 4 to both sides

x = 12/8 .... divide both sides by 8

x = 3/2 ... reduce

x = 1.5 .... convert to decimal form

Now plug x = 1.5 into the expression 2x-3 and simplify

2x-3 = 2*1.5-3 = 3-3 = 0 which is the answer

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The volume of triangular prims.

Nat2105 [25]

Check the picture below.

so we really have 2 equal triangles with 3 rectangles, so let's get the area of each and sum them up.

$\stackrel{\textit{two equal triangles}}{2\left[ \cfrac{1}{2}(8)(6) \right]}~~+~~\stackrel{back}{(6\cdot 4)}~~ +~~\stackrel{bottom}{(8\cdot 4)}~~+~~\stackrel{top}{(10\cdot 4)}\implies 144$

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2 years ago

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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3 years ago

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3 years ago

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3 years ago

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Butoxors [25]

Answer:

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Step-by-step explanation:

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You can leave the answer as is or round it to 3.1

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4 years ago