Quadratic Equations Practice Problems - Solve by 3 Methods

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.

The parameters are expressed in the quadratic equation as follows:

aX2+bX+c=0

We substitute into the formula:

-5±√(5²-4*1*4) 
          2

-5±√(25-16)
         2

-5±√9
    2

-5±3
   2

The symbol ± means that we have to solve this part twice, once with a plus and a second time with a minus,

This is how we later get two results.

-5-3 = -8
-8/2 = -4

-5+3 = -2
-2/2 = -1

And thus we find out that X = -1, -4