Question

If ${\log }_{10}(x^2-3x+6)=1$, the value of $x$ is

• A. $10$ or $2$
• B. $4$ or $-2$
• C. $3$ or $-1$
• D. $4$ or $-1$
• E. None of these

Answer 1

Answer: (D)

$4$ or $-1$

Given: ${\log }_{10}(x^2-3x+6)=1$

${\log }_{10}(x^2-3x+6)={\log }_{10}10$.......$\left(1\right)$ [since ${\log }_{a}a=1$]

From $\left(1\right)$:

$x^2-3x+6=10$

$\Rightarrow x^2-3x-4=0$

$\Rightarrow (x-4)(x+1)=0$

$\Rightarrow x=-1or4$

But we should also check that $f\left(x\right)=x^2-3x+6>0$ for $x=-1,4$........[for

${\log }_{a}N$; $N>0$]

$f(-1)=(-1{)}^2-3(-1)+6=10>0$

$f\left(4\right)=4^2-3\left(4\right)+6=10>0$

Hence both values are satisfying the condition.

So, $x=-1,4$