Find the solution(s) using the quadratic formula.32z2−54z−1=0\frac32z^2-\frac54z-1=0

z = 4 3 , z = − 1 2 z=\frac43,z=-\frac12

Find the solution(s) using the quadratic formula. 3 2 z 2 − 5 4 z − 1 = 0 \frac32z^2-\frac54z-1=0

For this problem, we're going to solve the quadratic three halves z squared minus five fourths z minus one equals zero using our quadratic formula. So first, let's identify our a value as three halves, our b value as negative five fourths, and our c value as negative one. And next, we're going to plug it into our quadratic formula. So though our formula usually says x equals here, since our variable is z, we're gonna go ahead and say z equals negative b, which is negative five fourths plus or minus the square root of b squared, so negative five fourths squared, minus four times a, which is three halves, times c, which is negative one, all divided by two times a, which is three halves. Simplifying, we get z is equal to negative negative five fourths, which is positive five fourths, Plus or minus our square root of negative five fourths square, which will make 25 sixteenths. Minus negative four times negative one will give us positive four. And positive four times three halves is 12 halves. All divided by two times three halves is three. Next, we need to combine these fractions. So to do that, we need to get a common denominator. So here we have 12 over two, but we need it to be something over 16. To find that something, we need to multiply by eight over eight because eight times two will give us sixteen, and eight times 12 then is 96. So I can rewrite this as 96 over 16, and we can further simplify and get five fourths plus or minus the square root of 25 sixteenths plus 96 sixteenths, which is a 121 sixteenths all over three. And now I can use our product property to separate these square roots. So I can think of this as the square root of one twenty one over the square root of 16. And we know that the square root of one twenty one is 11 and the square root of 16 is four. So I can rewrite all of this as z is equal to five fourths plus or minus 11 fourths all over three. And that gives us then two solutions. One where we have five fourths plus 11 fourths, which is 16 fourths all over three, which simplifies to z equals 16 divided by four is four. So four thirds is one of our solutions. Our second solution is going to be five fourths minus 11 fourths, which is negative six fourths all over three. Now here six fourths does not reduce to a whole number like 16 fourths did. So So instead, we're going to think of this as a fraction over a fraction, which we can simplify by flipping the second and multiplying. So we'll have negative six fourths times the reciprocal of three over one, which is one third. And these reduce to make negative two fourths and reduces to z equals negative one half as our second solution. And as always, we could take our solutions here, plug them back into our original quadratic to check that our equation is true.

Find the solution(s) using the quadratic formula. 3 2 z 2 − 5 4 z − 1 = 0 \frac32z^2-\frac54z-1=0

z = 4 3 , z = − 1 2 z=\frac43,z=-\frac12

z = 4 3 , z = − 1 6 z=\frac43,z=-\frac16

z = 5 4 , z = − 1 2 z=\frac54,z=-\frac12

z = 5 4 , z = − 1 6 z=\frac54,z=-\frac16

12. Quadratic Equations and Functions

The Quadratic Formula

Beginning & Intermediate Algebra

12. Quadratic Equations and Functions

The Quadratic Formula

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Multiple Choice

Find the solution(s) using the quadratic formula.
$\frac{3}{2} z^{2} - \frac{5}{4} z - 1 = 0$

$z = \frac{4}{3}, z = - \frac{1}{6}$

$z = \frac{4}{3}, z = - \frac{1}{2}$

$z = \frac{5}{4}, z = - \frac{1}{2}$

$z = \frac{5}{4}, z = - \frac{1}{6}$

Verified step by step guidance

Identify the coefficients from the quadratic equation $\frac{3}{2}z^2 - \frac{5}{4}z - 1 = 0$. Here, $a = \frac{3}{2}$, $b = -\frac{5}{4}$, and $c = -1$.

Write down the quadratic formula: $z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Substitute the values of $a$, $b$, and $c$ into the formula: $z = \frac{-\left(-\frac{5}{4}\right) \pm \sqrt{\left(-\frac{5}{4}\right)^2 - 4 \cdot \frac{3}{2} \cdot (-1)}}{2 \cdot \frac{3}{2}}$.

Simplify inside the square root (the discriminant): calculate $b^2 - 4ac = \left(-\frac{5}{4}\right)^2 - 4 \cdot \frac{3}{2} \cdot (-1)$.

Simplify the numerator and denominator separately, then write the two possible solutions for $z$ using the $\pm$ sign, representing the two roots of the quadratic equation.

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