Limits – Practice questions ( Qn 1 – 9)

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Sure, here’s a step-by-step solution to the quadratic equation \(x^2 – 5x + 6 = 0\):

1. **Identify coefficients**:
The quadratic equation is in the form \(ax^2 + bx + c = 0\), where \(a = 1\), \(b = -5\), and \(c = 6\).

2. **Use the quadratic formula**:
The quadratic formula is \(x = \frac{{-b \pm \sqrt{{b^2 – 4ac}}}}{{2a}}\). Plug in the values of \(a\), \(b\), and \(c\):

\[
x = \frac{{-(-5) \pm \sqrt{{(-5)^2 – 4 \cdot 1 \cdot 6}}}}{{2 \cdot 1}}
\]

3. **Calculate discriminant**:
The discriminant \(D\) is \(b^2 – 4ac\). In this case, \(D = (-5)^2 – 4 \cdot 1 \cdot 6\).

\[
D = 25 – 24 = 1
\]

4. **Determine the nature of roots**:
– If \(D > 0\), there are two distinct real roots.
– If \(D = 0\), there is one real root (a repeated root).
– If \(D < 0\), there are two complex roots.

Here, since \(D = 1\), we have two distinct real roots.

5. **Solve for \(x\)**:
Substitute the values of \(a\), \(b\), \(c\), and \(D\) into the quadratic formula:

\[
x = \frac{{5 \pm \sqrt{1}}}{{2}}
\]

6. **Simplify**:
Since \(\sqrt{1} = 1\), we have:

\[
x_1 = \frac{{5 + 1}}{{2}} = \frac{6}{2} = 3
\]
\[
x_2 = \frac{{5 – 1}}{{2}} = \frac{4}{2} = 2
\]

7. **Check solutions**:
Substitute the values of \(x_1\) and \(x_2\) back into the original equation to ensure they satisfy it.

For \(x = 3\):
\[
(3)^2 – 5(3) + 6 = 9 – 15 + 6 = 0
\]

For \(x = 2\):
\[
(2)^2 – 5(2) + 6 = 4 – 10 + 6 = 0
\]

Both solutions satisfy the equation.

So, the solutions to the equation \(x^2 – 5x + 6 = 0\) are \(x = 3\) and \(x = 2\).

Question No: 1      [ Maximum Mark: 7 ]       (Do not use GDC)