Solve a ^ 2 + 1/(a ^ 2) = 3 | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$a^{2}+\frac{1}{a^{2}}=3$$

Solve for a

$a=\frac{-\sqrt{5}-1}{2}\approx -1.618033989$
$a = \frac{\sqrt{5} + 1}{2} \approx 1.618033989$
$a=\frac{\sqrt{5}-1}{2}\approx 0.618033989$
$a=\frac{1-\sqrt{5}}{2}\approx -0.618033989$

Steps Using the Quadratic Formula

Variable $a$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $a^{2}$.

$$a^{2}a^{2}+1=3a^{2}$$

To multiply powers of the same base, add their exponents. Add $2$ and $2$ to get $4$.

$$a^{4}+1=3a^{2}$$

Subtract $3a^{2}$ from both sides.

$$a^{4}+1-3a^{2}=0$$

Substitute $t$ for $a^{2}$.

$$t^{2}-3t+1=0$$

All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. Substitute $1$ for $a$, $-3$ for $b$, and $1$ for $c$ in the quadratic formula.

$$t=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 1\times 1}}{2}$$

Do the calculations.

$$t=\frac{3±\sqrt{5}}{2}$$

Solve the equation $t=\frac{3±\sqrt{5}}{2}$ when $±$ is plus and when $±$ is minus.

$$t=\frac{\sqrt{5}+3}{2}$$ $$t=\frac{3-\sqrt{5}}{2}$$

Since $a=t^{2}$, the solutions are obtained by evaluating $a=±\sqrt{t}$ for each $t$.

$$a=\frac{\sqrt{5}+1}{2}$$ $$a=-\frac{\sqrt{5}+1}{2}$$ $$a=-\frac{1-\sqrt{5}}{2}$$ $$a=\frac{1-\sqrt{5}}{2}$$