Question

If one of the roots of $x^2-bx+c=0,(b,c)ϵQ$ is $\sqrt{7-4\sqrt{3}}$ then:

• A. ${\log }_{b}c=0$
• B. $b+c=5$
• C. ${\log }_{c}b=0$
• D. $bc=-4$

Answer 1

Answer: (A), (B)

The correct options are
A ${\log }_{b}c=0$
B $b+c=5$

Given equation $x^2-bx+c=0,(b,c)\in Q$

Also given that one of the roots is $\sqrt{7-4\sqrt{3}}$

Consider $\sqrt{7-4\sqrt{3}}$

Rewrite $7$ as $4+3$ we get

$\sqrt{7-4\sqrt{3}}=\sqrt{4+3-4\sqrt{3}}$

$=\sqrt{2^2+(\sqrt{3}{)}^2-2\cdot 4\cdot \sqrt{3}}$

$=\sqrt{(2+\sqrt{3}{)}^2}$

$=2+\sqrt{3}$

Thus one root of $x^2-bx+c=0$ is $2+\sqrt{3}$ which is an irrational root.

We know that, irrational root occurs in conjugate pairs. Therefore the other root of $x^2-bx+c=0$ is$2-\sqrt{3}$.

Let $\alpha =2+\sqrt{3},\beta =2-\sqrt{3}$

Thus $\alpha ,\beta$ are roots of $x^2-bx+c=0$.

We know that sum of the roots is $b$ and product of roots is $c$

$\Rightarrow \alpha +\beta =b,\alpha \times \beta =c$

$\Rightarrow b=2+\sqrt{3}+2-\sqrt{3},c=(2+\sqrt{3})\times (2-\sqrt{3})$

$\Rightarrow b=4,c=2^2-(\sqrt{3}{)}^2=4-3=1$

Thus we get $b=4,c=1$

Hence $b+c=4+1=5$

and $lo{g}_{b}c=lo{g}_{4}1=0$ (since $log1=0$)

Therefore if one of the roots of $x^2-bx+c=0,(b,c)\in Q$ is $\sqrt{7-4\sqrt{3}}$ then $lo{g}_{b}c=0$ and $b+c=5$