The Quadratic Formula: System of equations with no solution

Let's identify that this is a quadratic equation:

$2x^2-3x+5=0$

and this is because 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 with the solving technique:

We'll choose to solve it using the quadratic formula,

Let's recall it first:

The rule states that the roots of an equation in 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 get a true statement when substituted in the equation)

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

Let's return to the problem:

$2x^2-3x+5=0$and solve it:

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

$\begin{cases}a=2 \\ b=-3 \\ c=5\end{cases}$

where we noted that the coefficient includes the minus sign, and this is because in the general form of the equation we mentioned earlier:

$ax^2+bx+c=0$

the coefficients are defined such that they have a plus sign in front of them, and therefore the minus sign must be included in the coefficient value.

Let's continue and get the equation's solutions (roots) by substituting the coefficients we noted earlier in the quadratic formula:

$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-(-3)\pm\sqrt{(-3)^2-4\cdot2\cdot5}}{2\cdot2}$

Let's continue and calculate the expression under the root and simplify the expression:

$x_{1,2}=\frac{3\pm\sqrt{-31}}{2}\frac{}{}$

We got that the expression under the root is negative, and since we cannot extract a real root from a negative number, this equation has no real solutions,

Meaning - there is no real value of $x$that when substituted in the equation will give a true statement.

Therefore, the correct answer is answer D.

First, we'll identify that this is a quadratic equation:

$-x^2+6x-10=0$

and this is because 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 with the solving technique:

For ease of solving and minimizing errors, it is always recommended to ensure that the coefficient of the quadratic term in the equation is positive,

We'll achieve this by multiplying (both sides of) the equation by:$-1$:

$-x^2+6x-10=0 \hspace{8pt}\text{/}\cdot(-1)\\ x^2-6x+10=0$$$

Let's continue solving the equation:

We'll choose to solve it using the quadratic formula,

Let's recall it first:

The rule states that the roots of an 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 get a true statement when substituted in the equation)

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

Let's return to the problem:

$x^2-6x+10=0$

and solve it:

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

$\begin{cases}a=1 \\ b=-6 \\ c=10\end{cases}$

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

And we'll get the equation's solutions (roots) by substituting these coefficients that we mentioned earlier in the quadratic formula:

$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-(-6)\pm\sqrt{(-6)^2-4\cdot1\cdot10}}{2\cdot1}$

Let's continue and calculate the expression under the root and simplify the expression:

$x_{1,2}=\frac{6\pm\sqrt{-4}}{2}$

We got that the expression under the root is negative, and since we cannot extract a real root from a negative number, this equation has no real solutions,

Meaning - there is no real value of $x$that when substituted in the equation will give a true statement.

Therefore, the correct answer is answer D.

The parameters are expressed in the quadratic equation as follows:

aX2+bX+c=0

We identify that we have:
a=1
b=0
c=9

We recall the root formula:

We replace according to the formula:

-0 ± √(0²-4*1*9)

           2

We will focus on the part inside the square root (also called delta)

√(0-4*1*9)

√(0-36)

√-36

It is not possible to take the square root of a negative number.

And so the question has no solution.