Question

If one of the roots of $2x^2-cx+3=0$ is $3$ and another equation $2x^2-cx+d=0$ has equal roots, where $c\text{and}d$ are real numbers, then $d$ is equal to

• A. $3$
• B. $\frac{49}{8}$
• C. $\frac{8}{49}$
• D. $-3$

Answer 1

The correct option is B

$\frac{49}{8}$

Explanation for the correct option.

Step 1: Determinantion of the value of $\mathit{c}$.

Now the root of the equation $2x^2-cx+3=0$ ts $3$ then put $x=3$ in the equation,

$\begin{array}{rcl}\therefore 2{\left(3\right)}^2-3c+3& =& 0\\ & \Rightarrow & 18-3c+3=0\\ & \Rightarrow & c=7\end{array}$

Then the second equation with equal roots become,

$2x^2-7x+d=0$

Step 2: Determination of the value of $\mathit{d}$.

We know that when an equation have equal roots then its discriminant $\left(D\right)$ is equal to zero. And $D=B^2-4AC$, where $B=-7,A=2,\text{and}C=d$.

Applying the condition will give

$B^2-4AC=0$

This can be solved as,

$\begin{array}{rcl}\therefore {\left(-7\right)}^2-4\left(2\right)\left(d\right)& =& 0\\ & \Rightarrow & 8d=49\\ & \Rightarrow & d=\frac{49}{8}\end{array}$

Hence, the correct option is (B).