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Unproctored Placement Assessment

Solve for [tex]$u$[/tex].

[tex]
(u+3)^2 = 2u^2 + 16u + 30
[/tex]

If there is more than one solution, separate them with a comma.

[tex]
u = \square
[/tex]


Sagot :

Sure, let's solve the equation step by step.

We start with the given equation:
[tex]\[ (u + 3)^2 = 2u^2 + 16u + 30 \][/tex]

First, we'll expand the left side of the equation:
[tex]\[ (u + 3)^2 = u^2 + 6u + 9 \][/tex]

So the equation now looks like:
[tex]\[ u^2 + 6u + 9 = 2u^2 + 16u + 30 \][/tex]

Next, let's move all the terms to one side of the equation to set it to 0. We'll move the terms on the right to the left by subtracting them:
[tex]\[ u^2 + 6u + 9 - 2u^2 - 16u - 30 = 0 \][/tex]

Combine like terms:
[tex]\[ u^2 - 2u^2 + 6u - 16u + 9 - 30 = 0 \][/tex]

Simplify:
[tex]\[ -u^2 - 10u - 21 = 0 \][/tex]

To make the leading coefficient positive, we can multiply the entire equation by -1:
[tex]\[ u^2 + 10u + 21 = 0 \][/tex]

This is now a standard quadratic equation in the form [tex]\(au^2 + bu + c = 0\)[/tex]. We can solve for [tex]\(u\)[/tex] using the quadratic formula:
[tex]\[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]

Here, [tex]\(a = 1\)[/tex], [tex]\(b = 10\)[/tex], and [tex]\(c = 21\)[/tex]. Plugging these values into the quadratic formula, we get:
[tex]\[ u = \frac{-10 \pm \sqrt{10^2 - 4 \cdot 1 \cdot 21}}{2 \cdot 1} \][/tex]

Simplify inside the square root:
[tex]\[ u = \frac{-10 \pm \sqrt{100 - 84}}{2} \][/tex]
[tex]\[ u = \frac{-10 \pm \sqrt{16}}{2} \][/tex]
[tex]\[ u = \frac{-10 \pm 4}{2} \][/tex]

Now, solve for the two possible values of [tex]\(u\)[/tex]:

1. When using the [tex]\(+\)[/tex]:
[tex]\[ u = \frac{-10 + 4}{2} = \frac{-6}{2} = -3 \][/tex]

2. When using the [tex]\(-\)[/tex]:
[tex]\[ u = \frac{-10 - 4}{2} = \frac{-14}{2} = -7 \][/tex]

Therefore, the solutions to the equation are:
[tex]\[ u = -3 \quad \text{and} \quad u = -7 \][/tex]

So, the solutions are [tex]\(u = -3\)[/tex] and [tex]\(u = -7\)[/tex].