Question

which equation is quadratic in form?
○ $2(x + 5)^3 + 8x + 5 + 6 = 0$
○ $x^6 + 6x^4 + 8 = 0$
○ $7x^6 + 36x^3 + 5 = 0$
○ $4x^9 + 20x^3 + 25 = 0$

Explanation:

Step1: Recall Quadratic Form

A quadratic - form equation is of the form \(au^{2}+bu + c = 0\), where \(u\) is a function of \(x\), and the highest power of \(u\) is 2.

Step2: Analyze Option 1

For the equation \(2(x + 5)^{3}+8(x + 5)+6 = 0\) (assuming a typo, it should be \(8(x + 5)\) instead of \(8x + 5\)), if we let \(u=x + 5\), the equation becomes \(2u^{3}+8u + 6=0\). The highest power of \(u\) is 3, so it is not a quadratic - form equation.

Step3: Analyze Option 2

For the equation \(x^{6}+6x^{4}+8 = 0\), if we let \(u = x^{3}\), then \(u^{2}=x^{6}\), and the equation becomes \(u^{2}+6x^{4}+8 = 0\). But \(x^{4}=(x^{2})^{2}\), and if we let \(u=x^{2}\), then \(u^{3}=x^{6}\) and \(u^{2}=x^{4}\), the equation is \(u^{3}+6u^{2}+8 = 0\). The highest power of \(u\) is 3, so it is not a quadratic - form equation.

Step4: Analyze Option 3

For the equation \(7x^{6}+36x^{3}+5 = 0\), let \(u=x^{3}\). Then \(x^{6}=(x^{3})^{2}=u^{2}\). Substituting \(u = x^{3}\) into the equation, we get \(7u^{2}+36u + 5=0\), which is in the form \(au^{2}+bu + c = 0\) (where \(a = 7\), \(b = 36\), \(c = 5\) and \(u=x^{3}\)).

Step5: Analyze Option 4

For the equation \(4x^{9}+20x^{3}+25 = 0\), if we let \(u=x^{3}\), then \(x^{9}=(x^{3})^{3}=u^{3}\). The equation becomes \(4u^{3}+20u + 25=0\). The highest power of \(u\) is 3, so it is not a quadratic - form equation.

Answer:

\(7x^{6}+36x^{3}+5 = 0\) (the third option)