Question

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

• A. – 2 and 7
• B. – 2 and –7
• C. 2 and 7
• D. 2 and –7

Answer 1

The correct option is D 2 and –7

Given, One root is 5 and $\left|\begin{array}{ccc}x& 3& 7\\ 2& x& -2\\ 7& 8& x\end{array}\right|=0$
Expanding the given equation, we get
$x[x^2-(-16\left)\right]-3[2x-(-14\left)\right]+7[16-7x]=0\Rightarrow x^3+16x-6x-42+112-49x=0\Rightarrow x^3-39x+70=0$
Since 5 is the one root of given equation, therefore $x^3-5x^2+5x^2-25x-14x+70=0\Rightarrow x^2(x-5)+5x(x-5)-14(x-5)=0\Rightarrow (x-5)(x^2+5x-14)=0\Rightarrow (x-5)(x-2)(x+7)=0$ or x = 5, 2 and -7.