Question

which equation is quadratic in form?
○ $4(x - 2)^2 + 3x - 2 + 1 = 0$
○ $8x^5 + 4x^3 + 1 = 0$
○ $10x^8 + 7x^4 + 1 = 0$
○ $9x^{16} + 6x^4 + 1 = 0$

Explanation:

Step1: Recall quadratic form

A quadratic - form equation is of the form \(a(u)^2 + b(u)+c = 0\), where \(u\) is a function of \(x\) and the highest power of \(u\) is 2. In terms of polynomial equations, if we can substitute \(u = x^n\) such that the equation becomes \(au^{2}+bu + c=0\), then the equation is quadratic in form.

Step2: Analyze each option

• Option 1: \(4(x - 2)^{2}+3x - 2 + 1=0\). Expand \((x - 2)^{2}=x^{2}-4x + 4\). The equation becomes \(4(x^{2}-4x + 4)+3x-2 + 1=0\), which simplifies to \(4x^{2}-16x+16 + 3x-1 = 0\) or \(4x^{2}-13x + 15=0\). But this is a quadratic equation in \(x\), not in a substituted form. The original form is not a quadratic - form equation in the sense of substituting a power of \(x\) (it is already a quadratic in \(x\) after expansion).
• Option 2: \(8x^{5}+4x^{3}+1 = 0\). Let \(u=x^{3}\), then \(x^{5}=x^{2}\cdot x^{3}=x^{2}u\). The equation becomes \(8x^{2}u+4u + 1=0\), which is not a quadratic equation in \(u\) (because of the \(x^{2}\) term).
• Option 3: \(10x^{8}+7x^{4}+1 = 0\). Let \(u = x^{4}\). Then \(x^{8}=(x^{4})^{2}=u^{2}\). The equation becomes \(10u^{2}+7u + 1=0\), which is a quadratic equation in \(u\).
• Option 4: \(9x^{16}+6x^{4}+1 = 0\). Let \(u=x^{4}\), then \(x^{16}=(x^{4})^{4}=u^{4}\). The equation becomes \(9u^{4}+6u + 1=0\), which is a quartic (degree 4) equation in \(u\), not a quadratic equation.

Answer:

\(10x^{8}+7x^{4}+1 = 0\) (the third option)