Question

If the equation $2a{x}^2-3bx+4c=0$ and $3x^2-4x+5=0$have a common root, then $\frac{(a+b)}{(b+c)}$ is equal to $(a,b,c\in \mathrm{ℝ})$

• A. $\frac{1}{2}$
• B. $\frac{3}{35}$
• C. $\frac{34}{31}$
• D. $\frac{29}{23}$
• E. None of these

Answer 1

The correct option is C

$\frac{34}{31}$

Explanation for correct answer

Step 1: Solve for the coefficients

We know that is $\alpha ,\beta$ are the roots of a quadratic equation $p{x}^2+qx+r=0$ then

$\alpha +\beta =-\frac{q}{p}$

$\alpha \beta =\frac{r}{p}$

Given that the roots are common roots.

So the coefficient of both the quadratic equation will be equal

Now comparing we have $2a=3$, $3b=4$ and $4c=5$

Therefore $a=\frac{3}{2}$, $b=\frac{4}{3}$and $c=\frac{5}{4}$

Step 2: Solve for the value of the given expression

The given expression is $\frac{(a+b)}{(b+c)}$

$$\begin{gathered} =\frac{{\displaystyle \frac{3}{2}}+{\displaystyle \frac{4}{3}}}{{\displaystyle \frac{4}{3}}+{\displaystyle \frac{5}{4}}} \\ =\frac{{\displaystyle \frac{9+8}{6}}}{{\displaystyle \frac{16+15}{12}}} \\ =\frac{34}{31} \end{gathered}$$

Hence, option (C) is correct i.e. $\frac{34}{31}$