Solve the following quadratic equations.2x2=5x+32x^2=5x+3

x = − 1 2 x=-\frac12 x = − 2 1 or x = 3 x=3 x = 3

Solve the following quadratic equations. 2 x 2 = 5 x + 3 2x^2=5x+3

Here we've been asked to solve this quadratic equation and we're going to do this by factoring. So first, we need to write this equation in standard form and right now it's not. We need to get all the terms on one side. So to do that, we're going to subtract five x and three from both sides so that we have two x squared minus five x minus three equals zero. Now that it's in standard form, we can apply our factoring techniques. Since our coefficient in front of our x squared term is not one, we're going to apply the AC method. So for our AC method, we have a is equal to two and c is equal to negative three, which makes negative six, and our b value is negative five. So we want to find two numbers that multiply to negative six and add to negative five. So let's take a look at one and negative six. These do add to negative five and multiply to negative six, so this is the factor pair we're looking for. Next, we're going to split up our middle term as the sum of these factor pairs. So we can rewrite our equation as two x squared plus x plus negative six x minus three equals zero. And now we can apply factoring by grouping. So we're going to group our first two terms and our last two terms and look for a GCF. For our first two terms, we have a GCF of x and for our last two terms, we have a GCF of negative three. So factoring those out, we have x times two x plus one, plus negative three times two x plus one. And now we have common binomials as well that we can also factor out. Doing that we have two x plus one times x minus three equals zero. Now we're ready to apply our zero product property. So in order to solve this, I want to know when does two x plus one equal zero and when does x minus three equal zero. Well, solving for x in two x plus one, we subtract one from both sides. We have two x equals negative one, then divide by two, and we're left with x equals negative one half. And for x minus three, we add three to both sides and we get x equals three. So these are the two solutions to this equation and we could always check our work by taking these values and plugging them back into our original equation to make sure that it is true.

Solve the following quadratic equations. 2 x 2 = 5 x + 3 2x^2=5x+3

x = − 1 2 x=-\frac12 x = − 2 1 or x = 3 x=3 x = 3

x = 1 2 x=\frac12 x = 2 1 or x = − 3 x=-3 x = − 3

x = − 1 2 x=-\frac12 x = − 2 1 or x = − 3 x=-3 x = − 3

x = 1 2 x=\frac12 x = 2 1 or x = 3 x=3 x = 3

7. Factoring

Quadratic Equations & Applications

Intermediate Algebra

7. Factoring

Quadratic Equations & Applications

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Multiple Choice

Solve the following quadratic equations.
$2 x^{2} = 5 x + 3$

$x=\frac12$ or $x=-3$

$x=-\frac12$ or $x=3$

$x=-\frac12$ or $x=-3$

$x=\frac12$ or $x=3$

Verified step by step guidance

Rewrite the given equation so that all terms are on one side, setting the equation equal to zero. Starting with \$2x^2 = 5x + 3$, subtract $5x$ and $3$ from both sides to get $2x^2 - 5x - 3 = 0$.

Identify the coefficients in the quadratic equation $ax^2 + bx + c = 0$. Here, $a = 2$, $b = -5$, and $c = -3$.

Use the quadratic formula to solve for $x$: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Substitute the values of $a$, $b$, and $c$ into the formula.

Calculate the discriminant $\Delta = b^2 - 4ac$ to determine the nature of the roots. This step helps to know if the solutions are real and distinct, real and equal, or complex.

Simplify the expression under the square root and then simplify the entire formula to find the two possible values for $x$. These will be your solutions to the quadratic equation.

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