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Vlad [161]
3 years ago
9

Solve x^2+ 8x- 6=0 using the Quadratic Formula

Mathematics
2 answers:
CaHeK987 [17]3 years ago
6 0

Answer:

x=−4+√22 or x=−4−√22

Step-by-step explanation:

Substitute all values in the quadratic formula.

Solve.

Black_prince [1.1K]3 years ago
3 0

Answer:

\boxed {\sf x= {{ - 4\pm \sqrt{22} }}}

Step-by-step explanation:

The quadratic equation is used to find the roots of a quadratic. The formula is:

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

when  {ax^2 + bx + c = 0}

We are given the quadratic: x^2+8x-6=0

If we compare the given quadratic to the standard form of a quadratic, then:

a= 1\\b=8 \\c= -6

Substitute the values into the formula.

x = \frac{{ - 8\pm \sqrt {8^2 - 4(1)(-6)} }}{{2(1)}}

Solve inside the radical first.

Solve the exponent.

  • 8²= 8*8= 64

x = \frac{{ - 8\pm \sqrt {64 - 4(1)(-6)} }}{{2(1)}}

Multiply 4, 1, and -6.

  • 4(1)(-6)= 4(-6)= -24

x = \frac{{ - 8\pm \sqrt {64 - -24}}}{{2(1)}}

Add 64 and 24 (2 negative signs become a positive)

  • 64- -24 64+24=88

x = \frac{{ - 8\pm \sqrt {88}}}{{2(1)}}

Solve the denominator.

x = \frac{{ - 8\pm \sqrt {88}}}{{2}}

The radical can be simplified. 88 is divisible by a perfect square: 4

x= \frac{{ - 8\pm \sqrt {4}\sqrt{22} }}{{2}}

Take the square root of 4.

x= \frac{{ - 8\pm 2\sqrt{22} }}{{2}}

Divide by 2.

x= {{ - 4\pm \sqrt{22} }}

The roots are: x=0.690416  and x=−8.69042

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A box with a square base and open top must have a volume of 296352 c m 3 . We wish to find the dimensions of the box that minimi
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Answer:

  • Base Length of 84cm
  • Height of 42 cm.

Step-by-step explanation:

Given a box with a square base and an open top which must have a volume of 296352 cubic centimetre. We want to minimize the amount of material used.

Step 1:

Let the side length of the base =x

Let the height of the box =h

Since the box has a square base

Volume, V=x^2h=296352

h=\dfrac{296352}{x^2}

Surface Area of the box = Base Area + Area of 4 sides

A(x,h)=x^2+4xh\\$Substitute h=\dfrac{296352}{x^2}\\A(x)=x^2+4x\left(\dfrac{296352}{x^2}\right)\\A(x)=\dfrac{x^3+1185408}{x}

Step 2: Find the derivative of A(x)

If\:A(x)=\dfrac{x^3+1185408}{x}\\A'(x)=\dfrac{2x^3-1185408}{x^2}

Step 3: Set A'(x)=0 and solve for x

A'(x)=\dfrac{2x^3-1185408}{x^2}=0\\2x^3-1185408=0\\2x^3=1185408\\$Divide both sides by 2\\x^3=592704\\$Take the cube root of both sides\\x=\sqrt[3]{592704}\\x=84

Step 4: Verify that x=84 is a minimum value

We use the second derivative test

A''(x)=\dfrac{2x^3+2370816}{x^3}\\$When x=84$\\A''(x)=6

Since the second derivative is positive at x=84, then it is a minimum point.

Recall:

h=\dfrac{296352}{x^2}=\dfrac{296352}{84^2}=42

Therefore, the dimensions that minimizes the box surface area are:

  • Base Length of 84cm
  • Height of 42 cm.
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