Solve the Quadratic Equation: 2x²-3x+5=0 Step-by-Step

Question

Solve the following equation:

2x23x+5=0 2x^2-3x+5=0

Video Solution

Solution Steps

00:00 Find X
00:04 Identify the coefficients
00:15 Use the roots formula
00:39 Substitute appropriate values according to the given data and solve
01:06 Calculate the square and products
01:20 A root cannot be a negative number
01:32 Therefore there is no solution to the question

Step-by-Step Solution

Let's identify that this is a quadratic equation:

2x23x+5=0 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:

ax2+bx+c=0 ax^2+bx+c=0

are:

x1,2=b±b24ac2a 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:

2x23x+5=0 2x^2-3x+5=0 and solve it:

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

{a=2b=3c=5 \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:

ax2+bx+c=0 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:

x1,2=b±b24ac2a=(3)±(3)242522 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:

x1,2=3±312 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 x that when substituted in the equation will give a true statement.

Therefore, the correct answer is answer D.

Answer

No solution