Solve for $$ x $$ : $$ \begin{array}{l} ext { 2.1.1 } \quad p x+q x=a \ 2.1 .2 \quad 2 x^{2}-5 x+2=0\end{array} $$

Question

Solve for $$ x $$ : $$ \begin{array}{l}	ext { 2.1.1 } \quad p x+q x=a \ 2.1 .2 \quad 2 x^{2}-5 x+2=0\end{array} $$

UpStudy AI · Accepted Answer

To solve for $$ x $$ in the given equations, we will address each equation separately.

### Equation 2.1.1: $$ p x + q x = a $$

We can factor out $$ x $$ from the left side:

$$
(p + q)x = a
$$

Now, to solve for $$ x $$, we divide both sides by $$ p + q $$ (assuming $$ p + q \neq 0 $$):

$$
x = \frac{a}{p + q}
$$

### Equation 2.1.2: $$ 2x^2 - 5x + 2 = 0 $$

This is a quadratic equation, and we can solve it using the quadratic formula:

$$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$

Here, $$ a = 2 $$, $$ b = -5 $$, and $$ c = 2 $$. Plugging in these values:

1. Calculate the discriminant:

$$
b^2 - 4ac = (-5)^2 - 4 \cdot 2 \cdot 2 = 25 - 16 = 9
$$

2. Now apply the quadratic formula:

$$
x = \frac{-(-5) \pm \sqrt{9}}{2 \cdot 2} = \frac{5 \pm 3}{4}
$$

This gives us two possible solutions:

$$
x_1 = \frac{5 + 3}{4} = \frac{8}{4} = 2
$$

$$
x_2 = \frac{5 - 3}{4} = \frac{2}{4} = \frac{1}{2}
$$

### Summary of Solutions

- From Equation 2.1.1, we have:

$$
x = \frac{a}{p + q}
$$

- From Equation 2.1.2, the solutions are:

$$
x = 2 \quad \text{and} \quad x = \frac{1}{2}
$$

Thus, the solutions for $$ x $$ are:

$$
x = \frac{a}{p + q}, \quad x = 2, \quad x = \frac{1}{2}
$$
From Equation 2.1.1, $$ x = \frac{a}{p + q} $$. From Equation 2.1.2, $$ x = 2 $$ and $$ x = \frac{1}{2} $$.

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Solution

Equation 2.1.1:

Equation 2.1.2:

Summary of Solutions

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