QuestionDeduce the solutions of the following equations: i. $x^{4}-5 x^{2}-6=0$ ii. $x-5 \sqrt{x}-6=0$

Studdy Solution

STEP 1

1. We are given two separate equations to solve.
2. Each equation may require different algebraic techniques to find the solutions.

STEP 2

1. Solve the first equation $x^4 - 5x^2 - 6 = 0$ .
2. Solve the second equation $x - 5\sqrt{x} - 6 = 0$ .

STEP 3

Consider the first equation: $x^4 - 5x^2 - 6 = 0$ .
Notice that this is a quadratic in form if we let $y = x^2$ . Thus, rewrite the equation as:
$y^2 - 5y - 6 = 0$

STEP 4

Solve the quadratic equation $y^2 - 5y - 6 = 0$ using the quadratic formula:
$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
where $a = 1$ , $b = -5$ , and $c = -6$ .
$y = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(-6)}}{2(1)}$ $y = \frac{5 \pm \sqrt{25 + 24}}{2}$ $y = \frac{5 \pm \sqrt{49}}{2}$ $y = \frac{5 \pm 7}{2}$

STEP 5

Calculate the two possible values for $y$ :
$y = \frac{5 + 7}{2} = 6$ $y = \frac{5 - 7}{2} = -1$

STEP 6

Since $y = x^2$ , substitute back to find $x$ :
For $y = 6$ : $x^2 = 6$ $x = \pm \sqrt{6}$
For $y = -1$ : $x^2 = -1$
Since $x^2 = -1$ has no real solutions, we discard this case.

STEP 7

Consider the second equation: $x - 5\sqrt{x} - 6 = 0$ .
Let $y = \sqrt{x}$ . Then $x = y^2$ .
Substitute into the equation:
$y^2 - 5y - 6 = 0$
This is the same quadratic equation as in the first problem.

STEP 8

Using the solutions from the previous quadratic equation:
$y = 6 \quad \text{or} \quad y = -1$
Since $y = \sqrt{x}$ , $y$ must be non-negative. Thus, $y = -1$ is not valid.

STEP 9

For $y = 6$ : $\sqrt{x} = 6$ $x = 36$
The solutions are:
For equation i: $x = \pm \sqrt{6}$
For equation ii: $x = 36$

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