Find the solution(s) using the quadratic formula.32(x+5)2−4=x+1\frac32\left(x+5\right)^2-4=x+1

x = − 13 3 , x = − 5 x=-\frac{13}{3},x=-5 x = − 3 13 , x = − 5

Find the solution(s) using the quadratic formula. 3 2 ( x + 5 ) 2 − 4 = x + 1 \frac32\left(x+5\right)^2-4=x+1

In this problem, we want to solve the quadratic three halves times x plus five squared minus four equals x plus one using our quadratic formula. Now, clearly, this is not in standard form. So our first step before we can apply our quadratic formula is to foil these out, combine all of our like terms, and get this in standard form. So to do that, we're going to first foil out x plus five squared. We can rewrite that as x plus five times x plus five. Using our foil technique, we have x times x, which makes x squared. Next, x times five, which makes five x. Then five times x, which gets us another five x. And finally, five times five, which is 25. Combining like terms, we have x squared plus 10 x plus 25. So now I can take this quadratic and plug it back in for x plus five squared. So we'll have three halves times x squared plus 10 x plus 25 minus four equals x plus one. Now we can distribute in the three halves. So distributing this in, we get three halves x squared plus three times 10 is thirty, and thirty divided by two is 15. So we'll have plus 15 x. And then three times 25 is 75, and 75 divided by two is not a whole number. So I'm gonna go ahead and leave that as 75 halves. And that's all minus four and equal to x plus one. Now let's combine like terms. We have 75 halves and negative four. Now in order to combine these, we need to have a common denominator. So we can rewrite four as eight halves. So we have a three halves x squared plus 15 x, and 75 halves minus eight halves is 67 halves, and that's equal to x plus one. Now we want to get all of our terms on the same side, so we're going to subtract x and subtract one from both sides, And we're left with three halves x squared. 15 x minus x is 14 x. And 67 halves minus one, we can think of that as minus two halves. So 67 halves minus two halves is positive 65 halves, and this is now equal to zero. So now that we have our quadratic in standard form, we can identify our a value as three halves, our b value as 14, and our c value as 65 halves. And we can plug all of this into our quadratic formula. So we'll get x is equal to negative b, so negative 14, plus or minus the square root of b squared, so 14 squared, minus four times a, which is three halves, times c, which is 65 halves. And this is all over two times a, which is three halves. Simplifying, we can see that our two times three halves will reduce to three. And we can also see that here in our square root, we have two and two in the denominators, which will cancel out with our four out front here. So those will reduce to one, and we're left with negative 14 plus or minus the square root of 14 squared using our calculator is 196 minuteus three times 65, which using our calculator is 195. And that's all divided by three. Now, one ninety six minus one ninety five is just one, and the square root of one is one. So we can write this as x equals negative 14 plus or minus one, all divided by three. So we have two solutions here. The first is when x equals negative 14 plus one divided by three, which gives us negative 13 over three. And the second one is x equals negative 14 minus one divided by three. So negative 15 divided by three. Now negative 15 divided by three reduces to x equals negative five as one of our solutions. And negative 13 divided by three does not reduce, and so this is our second solution. We could always check our work by taking these solutions and plugging them back into our original equation to make sure that the equation is true.

Find the solution(s) using the quadratic formula. 3 2 ( x + 5 ) 2 − 4 = x + 1 \frac32\left(x+5\right)^2-4=x+1

x = − 13 3 , x = − 5 x=-\frac{13}{3},x=-5 x = − 3 13 , x = − 5

x = 13 x , x = − 5 x=\frac{13}{x},x=-5 x = x 13 , x = − 5

x = − 13 3 , x = 1 2 x=-\frac{13}{3},x=\frac12 x = − 3 13 , x = 2 1

x = − 5 , x = 5 x=-5,x=5 x = − 5 , x = 5

10. Quadratic Equations and Functions

The Quadratic Formula

Intermediate Algebra

10. Quadratic Equations and Functions

The Quadratic Formula

Struggling with Intermediate Algebra?

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Multiple Choice

Find the solution(s) using the quadratic formula.
$\frac{3}{2} {(x + 5)}^{2} - 4 = x + 1$

$x=$\frac{13}{x}$,x=-5$

$x=-$\frac{13}{3}$,x=-5$

$x=-$\frac{13}{3}$,x=$\frac$12$

$x=-5,x=5$

Verified step by step guidance

Start with the given equation: $\frac{3}{2}(x+5)^2 - 4 = x + 1$.

First, expand the squared term and simplify both sides to write the equation in standard quadratic form $ax^2 + bx + c = 0$.

Move all terms to one side of the equation to set it equal to zero.

Identify the coefficients $a$, $b$, and $c$ from the quadratic equation.

Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ to find the solutions for $x$.

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Struggling with Intermediate Algebra?

Find the solution(s) using the quadratic formula.32(x+5)2−4=x+1\(\frac\)32\(\left\)(x+5\(\right\))^2-4=x+1

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Find the solution(s) using the quadratic formula.
$\frac{3}{2} {(x + 5)}^{2} - 4 = x + 1$