Question 2

The equation [tex]$x^2 + 6x - 8 = 0$[/tex] has two solutions [tex]A[/tex] and [tex]B[/tex], where [tex]A \ \textless \ B[/tex].

[tex]A = \square[/tex] and [tex]B = \square[/tex]



Answer :

To solve the quadratic equation [tex]\(x^2 + 6x - 8 = 0\)[/tex] and find the solutions [tex]\(A\)[/tex] and [tex]\(B\)[/tex] such that [tex]\(A < B\)[/tex], we can follow these steps:

1. Identify the general form of the quadratic equation:
A quadratic equation generally has the form [tex]\(ax^2 + bx + c = 0\)[/tex]. In this case, the given equation is:
[tex]\[ x^2 + 6x - 8 = 0 \][/tex]
Here, [tex]\(a = 1\)[/tex], [tex]\(b = 6\)[/tex], and [tex]\(c = -8\)[/tex].

2. Use the quadratic formula to find the roots:
The quadratic formula, which solves the equation [tex]\(ax^2 + bx + c = 0\)[/tex], is given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substituting the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] into the quadratic formula yields:
[tex]\[ x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-6 \pm \sqrt{36 + 32}}{2} \][/tex]
[tex]\[ x = \frac{-6 \pm \sqrt{68}}{2} \][/tex]
[tex]\[ x = \frac{-6 \pm 2\sqrt{17}}{2} \][/tex]

3. Simplify the expression for the roots:
[tex]\[ x = -3 \pm \sqrt{17} \][/tex]
Therefore, the two solutions are:
[tex]\[ x_1 = -3 + \sqrt{17} \][/tex]
and
[tex]\[ x_2 = -3 - \sqrt{17} \][/tex]

4. Identify which solution is smaller and which is larger:
We need to determine which solution is [tex]\(A\)[/tex] (the smaller root) and which is [tex]\(B\)[/tex] (the larger root). Clearly:
[tex]\[ -3 - \sqrt{17} \][/tex]

Thus, the solutions are:
[tex]\[ A = -3 - \sqrt{17} \][/tex]
[tex]\[ B = -3 + \sqrt{17} \][/tex]

Therefore, the solutions are:
[tex]\[ A = -\sqrt{17} - 3 \quad \text{(the smaller root)} \][/tex]
[tex]\[ B = -3 + \sqrt{17} \quad \text{(the larger root)} \][/tex]

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