If 5 is one root of the equation left| | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(d) Given, One root =$ 5$ and equation $\left| {\,\begin{array}{*{20}{c}}x&3&7\2&x&{ - 2}\7&8&x\end{array}\,} ight|\, = 0$

Vedclass · Accepted Answer

$2 $ and $-7$

Vedclass · Answer

(d) Given, One root =$ 5$ and equation $\left| {\,\begin{array}{*{20}{c}}x&3&7\2&x&{ - 2}\7&8&x\end{array}\,} ight|\, = 0$.

Expanding the given equation, we get

$ + {\log _x}z({\log _y}x{\log _z}y - {\log _z}x)$

==> ${x^3} + 16x - 6x - 42 + 112 - 49x = 0$

==> ${x^3} - 39x + 70 = 0$

Since $ 5$  is the one root of given equation, therefore ${x^3} - 5{x^2} + 5{x^2} - 25x - 14x + 70 = 0$

==> ${x^2}(x - 5) + 5x(x - 5) - 14(x - 5) = 0$

==> $(x - 5)({x^2} + 5x - 14) = 0$

==> $(x - 5)\,(x - 2)\,(x + 7) = 0$ or $x = 5,\,2$ and $ -7.$

If $ 5$ is one root of the equation $\left| {\,\begin{array}{*{20}{c}}x&3&7\\2&x&{ - 2}\\7&8&x\end{array}\,} \right| = 0$, then other two roots of the equation are

Similar Questions

Evaluate the determinants

$\left|\begin{array}{ccc}0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0\end{array}\right|$

Evaluate the determinants

$\left|\begin{array}{ccc}2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0\end{array}\right|$

If system of equations $kx + 2y - z = 2,$$\left( {k - 1} \right)x + ky + z = 1,x + \left( {k - 1} \right)y + kz = 3$ has only one solution, then number of possible real value$(s)$ of $k$ is -

$\left| {\,\begin{array}{*{20}{c}}1&1&1\\1&{1 + x}&1\\1&1&{1 + y}\end{array}\,} \right| = $

Find area of the triangle with vertices at the point given in each of the following: $(1,0),(6,0),(4,3)$