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2 years ago

A student concluded that the solution to the equation sqrt2x+1 +3=0 is x=4. Do you agree? Why or why not?

Mathematics

1 answer:

Katyanochek1 [597]2 years ago

5 0

Just put 4 as x in the equation and solve it. The new equation is:

sqrt(2*4)+1+3 = 0

Now we can solve it.

sqrt(2*4) becomes sqrt8:

sqrt8+1+3=0. -------> sqrt8+4=0

The sqrt of 8 is about 2.83.. so 2.83+4 is:

6.83

So we have 6.83 as the answer

but 6.83 does not equal 0 so x cannot equal 4

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Select the two values of x that are roots of this equation.<br>x + 2x - 6 = 0

VLD [36.1K]

For this case we must solve the following quadratic equation:

$x ^ 2 + 2x-6 = 0$

Where:

$a = 1\\b = 2\\c = -6$

The solution is given by:

$x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}$

Substituting:

$x = \frac {-2 \pm \sqrt {2 ^ 2-4 (1) (- 6)}} {2 (1)}\\x = \frac {-2 \pm \sqrt {4 + 24}} {2}\\x = \frac {-2 \pm \sqrt {28}} {2}\\x = \frac {-2 \pm \sqrt {2 ^ 2 * 7}} {2}\\x = \frac {-2 \pm2 \sqrt {7}} {2}\\x = -1 \pm \sqrt {7}$

Thus, we have two roots:

$x_ {1} = - 1+ \sqrt {7}\\x_ {2} = - 1- \sqrt {7}$

ANswer:

$x_ {1} = - 1+ \sqrt {7}\\x_ {2} = - 1- \sqrt {7}$

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3 years ago

What is 31/10 as a mixed fraction

erma4kov [3.2K]

3 and 1 over 10 becasue 10 goes into 30 3 times

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2 years ago

NEED ASAP MATH!! 7TH GRADE

lozanna [386]

Answer:

Step-by-step explanation:

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2 years ago

If <img src="https://tex.z-dn.net/?f=x-12%5Csqrt%7Bx%7D%20%2B36%3D0" id="TexFormula1" title="x-12\sqrt{x} +36=0" alt="x-12\sqrt{

Elis [28]

Answer:

x = 36

Step-by-step explanation:

$x - 12\sqrt{x} + 36 = 0$

Subtract x and 36 from both sides.

$-12\sqrt{x} = -x - 36$

Divide both sides by -1.

$12\sqrt{x} = x + 36$

Square both sides.

$144x = x^2 + 72x + 1296$

Subtract 144x from both sides.

$0 = x^2 - 72x + 1296$

Factor the right side.

$0 = (x - 36)^2$

$x - 36 = 0$

$x = 36$

Since the solution of the equation involved squaring both sides, we musty check the answer for possible extraneous solutions.

Check x = 36:

$x - 12\sqrt{x} + 36 = 0$

$36 - 12\sqrt{36} + 36 = 0$

$36 - 12\times 6 + 36 = 0$

$36 - 72 + 36 = 0$

$0 = 0$

Since 0 = 0 is a true statement, the solution x = 36 is a valid solution.

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3 years ago

Fill in the blank.<br> (5x)2<br> a0x2

Yuki888 [10]

Answer: $a_0=25$

Step-by-step explanation:

It is given that, $(5x)^2=a_0x^2$

We need to find the value of $a_0$

As $(5x)^2=a_0x^2$

Solving LHS

$25x^2=a_0x^2$

Now comparing the coefficients of $x^2$

In LHS the coefficient of $x^2$ is 25

In RHS the coefficient of $x^2$ is $a_0$

It implies that, $a_0=25$

So, the value of $a_0$ is 25.

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2 years ago