Question

[Hà Nam Phan Đình]

Giải phương trình sau : \(3x^2+4x+10=2\sqrt{14x^2-7}\)

Answer 1

[Hồ Đại Việt]

\(3x^2+4x+10=2\sqrt{14x^2-7}\)

\(\Leftrightarrow2\sqrt{14x^2-7}=3x^2+4x+10\)

\(\Leftrightarrow\left(2\sqrt{14x^2-7}\right)^2=\left(3x^2+4x+10\right)^2\)

\(\Leftrightarrow56x^2-28=9x^4+76x^2+10+24x^3+80x\)\(\Leftrightarrow56x^2-28-9x^4-76x^2-100-24x^3-80x=0\)\(\Leftrightarrow-20x^2-128-9x^4-24x^3-80x=0\)

\(\Leftrightarrow-9x^4-18x^3-6x^3-12x^2-8x^2-16x-64x-128=0\)\(\Leftrightarrow-9x^3\cdot\left(x+2\right)-6x^2\cdot\left(x+2\right)-8x\cdot\left(x+2\right)-64\left(x+2\right)=0\)\(\Leftrightarrow-\left(x+2\right)\cdot\left(9x^3+18x^2+12x^2-24x+32x+64\right)=0\)\(\Leftrightarrow-\left(x+2\right)\cdot\left(9x^2\left(x+2\right)-12x\cdot\left(x+2\right)+32\left(x+2\right)\right)=0\)

\(\Leftrightarrow-\left(x+2\right)^2\cdot\left(9x^2-12x+32\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-\left(x+2\right)^2=0\\9x^2-12x+32=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x\notin R\end{matrix}\right.\)

\(\Leftrightarrow x=-2\)