Westonci.ca is your go-to source for answers, with a community ready to provide accurate and timely information. Get quick and reliable solutions to your questions from a community of experienced experts on our platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

The equation whose roots are numerically equal but opposite in sign of the roots of [tex]$2x^2 + 3x + 4 = 0$[/tex] is:

Sagot :

To find the equation whose roots are numerically equal but opposite in sign of the roots of [tex]\(2 x^2 + 3 x + 4 = 0\)[/tex], let's follow these detailed steps:

### Step 1: Find the roots of the original equation

The given quadratic equation is:

[tex]\[2 x^2 + 3 x + 4 = 0\][/tex]

Using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 2\)[/tex], [tex]\(b = 3\)[/tex], and [tex]\(c = 4\)[/tex]:

1. Calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac = 3^2 - 4 \cdot 2 \cdot 4 = 9 - 32 = -23 \][/tex]

Since the discriminant is negative, the equation has two complex roots.

2. Find the roots:
[tex]\[ x = \frac{-3 \pm \sqrt{-23}}{2 \cdot 2} = \frac{-3 \pm \sqrt{23}i}{4} \][/tex]

Thus, the roots of the equation [tex]\(2 x^2 + 3 x + 4 = 0\)[/tex] are:
[tex]\[ x_1 = \frac{-3 - \sqrt{23}i}{4}, \quad x_2 = \frac{-3 + \sqrt{23}i}{4} \][/tex]

### Step 2: Determine the roots of the new equation

The roots of the new equation need to be numerically equal but opposite in sign to the roots of the original equation. Therefore, the new roots are:
[tex]\[ x_1' = \frac{3 + \sqrt{23}i}{4}, \quad x_2' = \frac{3 - \sqrt{23}i}{4} \][/tex]

### Step 3: Form the new equation

To form a quadratic equation given its roots [tex]\(x_1'\)[/tex] and [tex]\(x_2'\)[/tex], we use the fact that if [tex]\(r_1\)[/tex] and [tex]\(r_2\)[/tex] are the roots, the quadratic equation can be written as:
[tex]\[x^2 - (r_1 + r_2)x + r_1 r_2 = 0\][/tex]

For the new roots [tex]\(x_1' = \frac{3 + \sqrt{23}i}{4}\)[/tex] and [tex]\(x_2' = \frac{3 - \sqrt{23}i}{4}\)[/tex]:

1. Sum of the roots [tex]\((x_1' + x_2')\)[/tex]:
[tex]\[ \frac{3 + \sqrt{23}i}{4} + \frac{3 - \sqrt{23}i}{4} = \frac{3 + 3}{4} = \frac{6}{4} = \frac{3}{2} \][/tex]

2. Product of the roots [tex]\((x_1' \cdot x_2')\)[/tex]:
[tex]\[ \left(\frac{3 + \sqrt{23}i}{4}\right) \left(\frac{3 - \sqrt{23}i}{4}\right) = \frac{(3 + \sqrt{23}i)(3 - \sqrt{23}i)}{16} = \frac{9 - 23i^2}{16} = \frac{9 - (-23)}{16} = \frac{9 + 23}{16} = \frac{32}{16} = 2 \][/tex]

### Step 4: Write the new quadratic equation

Using the sum and product of the roots, the new quadratic equation is:
[tex]\[ x^2 - \left(\frac{3}{2}\right)x + 2 = 0 \][/tex]

Therefore, the equation whose roots are numerically equal but opposite in sign of the roots of [tex]\(2 x^2 + 3 x + 4 = 0\)[/tex] is:

[tex]\[ x^2 - \frac{3}{2} x + 2 = 0 \][/tex]
We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.