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mr_godi [17]
3 years ago
9

7x^2-7=0 solve each equation by factoring

Mathematics
2 answers:
fredd [130]3 years ago
4 0
7x^2 - 7 = 0
7(x^2 - 1) = 0
7(x + 1)(x - 1) = 0
x = 1; x = -1
Usimov [2.4K]3 years ago
3 0
(1,0) (-1,0)
or if you prefer this format: x= 1,-1
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well if you cuanculated i think it might be 157.5

Step-by-step explanation:

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PLZ HELP ASAP!!! A rectangular prism with a volume of 4 cubic units is filled with cubes with side lengths of 1/3 unit. How many
VashaNatasha [74]

Answer:

<em>Hello There the Correct answer is 192.</em>

<h3> Explanation</h3>

Each cube with side lengths of 1/4 have a volume of (1/4)3 which means each cube's volume is 0.015625 cubic units. To find how many cubes are needed to fill the prism, divide 3 cubic units by 0.015625 cubic units. Your answer will be 192 cubes.

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Find the solutions to 2x⁴-24x²+40=0 and the x-intercepts of the graph of y=2x⁴-24x²+40.
jekas [21]

Answer:

The solutions and the x-intercepts of the polynomial 2x^4-24x^2+40 are:

x=\sqrt{10},\:x=-\sqrt{10},\:x=\sqrt{2},\:x=-\sqrt{2}

Step-by-step explanation:

Given a function <em>f</em> a solution or a root of <em>f</em> is a value x_{0} at which f(x_0)=0.

An x-intercept is a point on the graph where y is zero.

To find the solutions of the polynomial and the x-intercepts 2x^4-24x^2+40 you need to:

First, we need to factor the polynomial expression

Factor the common term

{\left(2 x^{4} - 24 x^{2} + 40\right)} = {\left(2 \left(x^{4} - 12 x^{2} + 20\right)\right)}

We can treat x^{4} - 12 x^{2} + 20 as a quadratic function with respect to x^2

Let u=x^2. We can rewrite x^{4} - 12 x^{2} + 20 in terms of u as follows:

u^2-12u+20

We need to solve the quadratic equation

u^2-12u+20=0

for this we can use the Quadratic Equation Formula:

For a quadratic equation of the form ax^2+bx+c=0 the solutions are

x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

\mathrm{For\:}\quad a=1,\:b=-12,\:c=20:\quad u_{1,\:2}=\frac{-\left(-12\right)\pm \sqrt{\left(-12\right)^2-4\cdot \:1\cdot \:20}}{2\cdot \:1}

u_1=\frac{-\left(-12\right)+\sqrt{\left(-12\right)^2-4\cdot \:1\cdot \:20}}{2\cdot \:1}\\u_1=10

u_2=\frac{-\left(-12\right)-\sqrt{\left(-12\right)^2-4\cdot \:1\cdot \:20}}{2\cdot \:1}\\u_2=2

the solutions to the quadratic equation are:

u=10,\:u=2

Therefore, u^2-12u+20=(u-10)(u-2)

Recall that u=x^2 so

2 x^{4} - 24 x^{2} + 40=2 \left(x^{2} - 10\right) \left(x^{2} - 2\right)=0

Using the Zero factor Theorem: If ab = 0, then either a = 0 or b = 0, or both a and b are 0.

x^2-10=0 roots are x_1=\sqrt{10}; x_2=-\sqrt{10}

x^{2} - 2=0 roots are x_1=\sqrt{2}; x_2=-\sqrt{2}

The solutions and the x-intercepts are:

x=\sqrt{10},\:x=-\sqrt{10},\:x=\sqrt{2},\:x=-\sqrt{2}

Because all roots are real roots the x-intercepts and the solutions are equal.

7 0
3 years ago
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