Certainly! Let's break down the calculations for both the general quadratic equation and the specific quadratic equation step by step.
### 1. General Quadratic Equation [tex]\( p x^2 + q x + r = 0 \)[/tex]
For a quadratic equation of the form [tex]\( p x^2 + q x + r = 0 \)[/tex] where [tex]\( p \neq 0 \)[/tex], you can find the roots using the quadratic formula:
[tex]\[
x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}
\][/tex]
In this case, [tex]\( a = p \)[/tex], [tex]\( b = q \)[/tex], and [tex]\( c = r \)[/tex]. Substituting these into the formula gives:
[tex]\[
x = \frac{{-q \pm \sqrt{{q^2 - 4pr}}}}{2p}
\][/tex]
Therefore, the roots of the general quadratic equation [tex]\( p x^2 + q x + r = 0 \)[/tex] are:
[tex]\[
x_1 = \frac{{-q - \sqrt{{q^2 - 4pr}}}}{2p}
\][/tex]
[tex]\[
x_2 = \frac{{-q + \sqrt{{q^2 - 4pr}}}}{2p}
\][/tex]
### 2. Specific Quadratic Equation [tex]\( x^2 - 25 = 0 \)[/tex]
Next, consider the specific quadratic equation [tex]\( x^2 - 25 = 0 \)[/tex]. This is simplified as:
[tex]\[
x^2 = 25
\][/tex]
Taking the square root of both sides, we obtain:
[tex]\[
x = \pm \sqrt{25}
\][/tex]
Since [tex]\( \sqrt{25} = 5 \)[/tex], the equation yields two solutions:
[tex]\[
x = 5 \quad \text{and} \quad x = -5
\][/tex]
Therefore, the roots of the quadratic equation [tex]\( x^2 - 25 = 0 \)[/tex] are:
[tex]\[
x_1 = 5 \quad \text{and} \quad x_2 = -5
\][/tex]
### Summary
- The roots of the general quadratic equation [tex]\( p x^2 + q x + r = 0 \)[/tex] are:
[tex]\[
x_1 = \frac{{-q - \sqrt{{q^2 - 4pr}}}}{2p} \quad \text{and} \quad x_2 = \frac{{-q + \sqrt{{q^2 - 4pr}}}}{2p}
\][/tex]
- The roots of the specific quadratic equation [tex]\( x^2 - 25 = 0 \)[/tex] are:
[tex]\[
x_1 = 5 \quad \text{and} \quad x_2 = -5
\][/tex]
