Question

1. $4x^4 + 35x^2 - 9 = 0$

Explanation:

Step1: Substitute $u=x^2$

Let $u = x^2$, so the equation becomes:
$$4u^2 + 35u - 9 = 0$$

Step2: Solve quadratic for $u$

Use quadratic formula $u=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$, where $a=4$, $b=35$, $c=-9$:
$$u=\frac{-35\pm\sqrt{35^2-4\times4\times(-9)}}{2\times4}$$
$$u=\frac{-35\pm\sqrt{1225+144}}{8}$$
$$u=\frac{-35\pm\sqrt{1369}}{8}$$
$$u=\frac{-35\pm37}{8}$$

Step3: Calculate valid $u$ values

Compute two solutions:
$u_1=\frac{-35+37}{8}=\frac{2}{8}=\frac{1}{4}$
$u_2=\frac{-35-37}{8}=\frac{-72}{8}=-9$
Since $u=x^2\geq0$, discard $u=-9$.

Step4: Solve for $x$ from $u=\frac{1}{4}$

Substitute back $u=x^2$:
$$x^2=\frac{1}{4}$$
$$x=\pm\sqrt{\frac{1}{4}}=\pm\frac{1}{2}$$

Answer:

$x=\frac{1}{2}$ and $x=-\frac{1}{2}$