Answer :
To solve the equation [tex]\(2x^2 - 2x - 35 = x^2\)[/tex], follow these steps:
1. Move all terms to one side of the equation:
Start by subtracting [tex]\(x^2\)[/tex] from both sides to get everything on one side, forming a standard quadratic equation.
[tex]\[
2x^2 - 2x - 35 - x^2 = 0
\][/tex]
2. Simplify the equation:
Combine like terms:
[tex]\[
(2x^2 - x^2) - 2x - 35 = 0
\][/tex]
This simplifies to:
[tex]\[
x^2 - 2x - 35 = 0
\][/tex]
3. Identify the coefficients:
In the quadratic equation [tex]\(x^2 - 2x - 35 = 0\)[/tex], the coefficients are:
- [tex]\(a = 1\)[/tex]
- [tex]\(b = -2\)[/tex]
- [tex]\(c = -35\)[/tex]
4. Calculate the discriminant:
The discriminant ([tex]\(D\)[/tex]) of a quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex] is given by:
[tex]\[
D = b^2 - 4ac
\][/tex]
Plugging in the values:
[tex]\[
D = (-2)^2 - 4 \times 1 \times (-35) = 4 + 140 = 144
\][/tex]
5. Solve for the roots using the quadratic formula:
The quadratic formula is:
[tex]\[
x = \frac{-b \pm \sqrt{D}}{2a}
\][/tex]
Since the discriminant ([tex]\(D\)[/tex]) is 144, a perfect square, we'll have two real and distinct solutions.
- First solution ([tex]\(x_1\)[/tex]):
[tex]\[
x_1 = \frac{-(-2) + \sqrt{144}}{2 \times 1} = \frac{2 + 12}{2} = \frac{14}{2} = 7.0
\][/tex]
- Second solution ([tex]\(x_2\)[/tex]):
[tex]\[
x_2 = \frac{-(-2) - \sqrt{144}}{2 \times 1} = \frac{2 - 12}{2} = \frac{-10}{2} = -5.0
\][/tex]
6. Conclusion:
The solutions to the quadratic equation [tex]\(2x^2 - 2x - 35 = x^2\)[/tex] are [tex]\(x = 7.0\)[/tex] and [tex]\(x = -5.0\)[/tex].
1. Move all terms to one side of the equation:
Start by subtracting [tex]\(x^2\)[/tex] from both sides to get everything on one side, forming a standard quadratic equation.
[tex]\[
2x^2 - 2x - 35 - x^2 = 0
\][/tex]
2. Simplify the equation:
Combine like terms:
[tex]\[
(2x^2 - x^2) - 2x - 35 = 0
\][/tex]
This simplifies to:
[tex]\[
x^2 - 2x - 35 = 0
\][/tex]
3. Identify the coefficients:
In the quadratic equation [tex]\(x^2 - 2x - 35 = 0\)[/tex], the coefficients are:
- [tex]\(a = 1\)[/tex]
- [tex]\(b = -2\)[/tex]
- [tex]\(c = -35\)[/tex]
4. Calculate the discriminant:
The discriminant ([tex]\(D\)[/tex]) of a quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex] is given by:
[tex]\[
D = b^2 - 4ac
\][/tex]
Plugging in the values:
[tex]\[
D = (-2)^2 - 4 \times 1 \times (-35) = 4 + 140 = 144
\][/tex]
5. Solve for the roots using the quadratic formula:
The quadratic formula is:
[tex]\[
x = \frac{-b \pm \sqrt{D}}{2a}
\][/tex]
Since the discriminant ([tex]\(D\)[/tex]) is 144, a perfect square, we'll have two real and distinct solutions.
- First solution ([tex]\(x_1\)[/tex]):
[tex]\[
x_1 = \frac{-(-2) + \sqrt{144}}{2 \times 1} = \frac{2 + 12}{2} = \frac{14}{2} = 7.0
\][/tex]
- Second solution ([tex]\(x_2\)[/tex]):
[tex]\[
x_2 = \frac{-(-2) - \sqrt{144}}{2 \times 1} = \frac{2 - 12}{2} = \frac{-10}{2} = -5.0
\][/tex]
6. Conclusion:
The solutions to the quadratic equation [tex]\(2x^2 - 2x - 35 = x^2\)[/tex] are [tex]\(x = 7.0\)[/tex] and [tex]\(x = -5.0\)[/tex].
