Quadratic Formula Practice Problems - Step-by-Step Solutions

Let's recall the quadratic formula:

We'll substitute the given data into the formula:

$x={{-(-10)\pm\sqrt{-10^2-4\cdot2\cdot(-12)}\over 2\cdot2}}$

Let's simplify and solve the part under the square root:

$x={{10\pm\sqrt{100+96}\over 4}}$

$x={{10\pm\sqrt{196}\over 4}}$

$x={{10\pm14\over 4}}$

Now we'll solve using both methods, once with the addition sign and once with the subtraction sign:

$x={{10+14\over 4}} = {24\over4}=6$

$x={{10-14\over 4}} = {-4\over4}=-1$

We've arrived at the solution: X=6,-1

Let's solve the problem step-by-step:

First, consider the given equation:

$x^2 + 7x = 0$

This equation is almost in the standard form of a quadratic equation:

$ax^2 + bx + c = 0$

Where:

• $a$ is the coefficient of $x^2$
• $b$ is the coefficient of $x$
• $c$ is the constant term (independent number)

Let's identify each of these components from the given equation:

• For $a$: The term with $x^2$ is $x^2$. In this case, the coefficient is implicitly 1, so $a = 1$.
• For $b$: The term with $x$ is $7x$. The coefficient of $x$ is 7, so $b = 7$.
• For $c$: There is no independent constant term visible, so we assume $c = 0$.

Thus, the components of the quadratic equation are:

$a = 1$, $b = 7$, $c = 0$

The correct choice from the provided options is : $a=1$, $b=7$, $c=0$

Let's solve this problem step-by-step by identifying the coefficients of the quadratic equation:

First, examine the given equation:

$5 - 6x^2 + 12x = 0$

To make it easier to identify the coefficients, we rewrite the equation in the standard quadratic form:

$-6x^2 + 12x + 5 = 0$

In this expression, we can now directly identify the coefficients:

• The coefficient of $x^2$ (quadratic term) is $a = -6$.
• The coefficient of $x$ (linear term) is $b = 12$.
• The constant term (independent number) is $c = 5$.

Thus, the components of the quadratic equation are:

$a = -6$, $b = 12$, $c = 5$

By comparing these values to the multiple-choice options, we can determine that the correct choice is:

Choice 4: $a = -6$, $b = 12$, $c = 5$

Therefore, the final solution is:

$a = -6$, $b = 12$, $c = 5$.

Let's recall the general form of the quadratic function:

$y=ax^2+bx+c$ The function given in the problem is:

$y=-x^2+25x$$c$is the free term (meaning the coefficient of the term with power 0),

In the function in the problem there is no free term,

Therefore, we can identify that:

$c=0$Therefore, the correct answer is answer A.

This is a quadratic equation:

$x^2+3x-18=0$

This is due to the fact that there is a quadratic term (meaning raised to the second power),

The first step in solving a quadratic equation is always arranging it in a form where all terms on one side are ordered from highest to lowest power (in descending order from left to right) and 0 on the other side,

Then we can choose whether to solve it using the quadratic formula or by factoring/completing the square.

The equation in the problem is already arranged, so let's proceed to solve it using the quadratic formula,

Remember:

The rule states that the roots of the equation of the form:

$ax^2+bx+c=0$

are:

$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

(meaning its solutions, the two possible values of the unknown for which we obtain a true statement when inserted into the equation)

$$This formula is called: "The Quadratic Formula"

Let's return to the problem:

$x^2+3x-18=0$

And solve it:

First, let's identify the coefficients of the terms:

$\begin{cases}a=1\\b=3\\c=-18\end{cases}$

where we noted that the coefficient of the quadratic term is 1,

And we'll obtain the solutions of the equation (its roots) by substituting the coefficients we just noted in the quadratic formula:

$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-3\pm\sqrt{3^2-4\cdot1\cdot(-18)}}{2\cdot1}$

Let's continue and calculate the expression inside the square root and simplify the expression:

$x_{1,2}=\frac{-3\pm\sqrt{81}}{2}=\frac{-3\pm9}{2}$

Therefore the solutions of the equation are:

$$$\begin{cases}x_1=\frac{-3+9}{2}=3 \\ x_2=\frac{-3-9}{2}=-6\end{cases}$

Therefore the correct answer is answer C.