What are the roots of the quadratic equation [tex]\( px^2 + qx + r = 0 \; (a \neq 0) \)[/tex]?

What are the roots of the quadratic equation [tex]\( x^2 - 25 = 0 \)[/tex]?



Answer :

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]