Practicing Success
Given that $b^2-ac <0, a > 0$. The value of $Δ=\begin{vmatrix}a&b&ax+by\\b&c&bx+cy\\ax+by&bx+cy&0\end{vmatrix}$, is |
zero positive negative $b^2 + ac$ |
negative |
We have, $Δ=\begin{vmatrix}a&b&ax+by\\b&c&bx+cy\\ax+by&bx+cy&0\end{vmatrix}$ $⇒Δ=\begin{vmatrix}a&b&ax+by\\b&c&bx+cy\\0&0&-(ax^2+2bxy+cy^2)\end{vmatrix}$ [Applying $R_3 → R_3-xR_1-yR_2$] $⇒Δ=(b^2-ac) (ax^2 + 2bxy + cy^2)$ Now, $b^2-ac <0$ and $a > 0$ ⇒ Discriminant of $ax^2 + 2bxy + cy^2$ is negative and $a > 0$. $⇒ax^2+2bxy+cy^2 > 0$ for all $x, y ∈ R$. $⇒ Δ=(b^2-ac) (ax^2 + 2bxy + cy^2) <0$ |