Quadratic Equations Practice Problems - Solve by 3 Methods

To answer the question, we'll need to recall the quadratic formula:

$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$

Let's remember that:

a is the coefficient of X²

b is the coefficient of X

c is the free term

And if we look again at the formula given to us:

a=1

b=10

c=9

Let's substitute into the formula:

$x = {-10 \pm \sqrt{10^2-4\cdot 1 \cdot 9} \over 2\cdot 1}$

Let's start by solving what's under the square root:

$x = {-10 \pm \sqrt{100-36} \over 2}$

$x = {-10 \pm \sqrt{64} \over 2}$

$x = {-10 \pm 8 \over 2}$

Now we'll solve twice, once with plus and once with minus

$x = {-10 +8 \over 2}= {-2 \over 2} = -1$

$x = {-10 -8 \over 2} = {-18 \over 2} =-9$

And we can see that we got two solutions, X=-1 and X=-9

And that's the solution!

To solve the quadratic equation $-2X^2 + 6X + 8 = 0$ using the quadratic formula, follow these steps:

• Step 1: Calculate the discriminant, $b^2 - 4ac$.

Here, $a = -2$, $b = 6$, and $c = 8$. Plug these into the formula: $b^2 - 4ac = 6^2 - 4(-2)(8) = 36 + 64 = 100$ Since the discriminant is greater than zero, the roots are real and distinct.

• Step 2: Apply the quadratic formula, $X = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Substituting the values, we have: $X = \frac{-6 \pm \sqrt{100}}{2(-2)}$ Simplifying inside the square root gives us: $X = \frac{-6 \pm 10}{-4}$ This leads to two possible solutions: - First, calculate with the positive square root: $X_1 = \frac{-6 + 10}{-4} = \frac{4}{-4} = -1$ - Second, calculate with the negative square root: $X_2 = \frac{-6 - 10}{-4} = \frac{-16}{-4} = 4$

Thus, the solutions to the equation are $X_1 = 4$ and $X_2 = -1$.

Verifying against the choices, the correct choice is: : $X_1=4, X_2=-1$.

Therefore, the solution is $X_1 = 4, X_2 = -1$.

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

To solve the quadratic equation $x^2 + 9x + 8 = 0$, we will use the factoring method because it appears simple to factor.

First, we attempt to factor the quadratic expression $x^2 + 9x + 8$. We look for two numbers that multiply to 8 (the constant term) and add up to 9 (the coefficient of the $x$ term).

These numbers are 1 and 8. So, we can write:

$x^2 + 9x + 8 = (x + 1)(x + 8) = 0$

Now, to find the solutions, we set each factor equal to zero:

1. $x + 1 = 0$ → $x = -1$
2. $x + 8 = 0$ → $x = -8$

Therefore, the solutions to the equation are $x_1 = -1$ and $x_2 = -8$.

Upon reviewing the multiple-choice answers, we find that the correct choice is the one that matches our solutions:

$x_1=-1$ $x_2=-8$

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