Question

If roots of $4{{x}^{2}}+5k=\left( 5k+1 \right)x$ differ by unity, then find the value of $k$?

Answer 1

Hint: We know that in a quadratic equation $a{{x}^{2}}+bx+c=0$, with roots $\alpha \text{ and }\beta $, $\alpha +\beta =-\dfrac{b}{a}$ and $\alpha \beta =\dfrac{c}{a}$. We can use these two values to find $\left( \alpha -\beta \right)$ and then equate it with 1, as given in the question. On further solving, we can calculate the required values of $k$.

Complete step by step solution:
We know that for a quadratic equation $a{{x}^{2}}+bx+c=0$, with roots $\alpha \text{ and }\beta $, the sum of roots $\alpha +\beta =-\dfrac{b}{a}$ and the product of roots $\alpha \beta =\dfrac{c}{a}$.
Here, in this question, we are given the quadratic equation $4{{x}^{2}}+5k=\left( 5k+1 \right)x$ which could also be written as
$4{{x}^{2}}-\left( 5k+1 \right)x+5k=0...\left( i \right)$
Let us assume the roots of this quadratic equation to be $\alpha \text{ and }\beta $.
So, by using the formulae of sum of roots and product of roots, we get
\[\alpha +\beta =-\dfrac{-\left( 5k+1 \right)}{4}\]
which could also be written as
\[\alpha +\beta =\dfrac{5k+1}{4}...\left( ii \right)\]
and the product of roots as
$\alpha \beta =\dfrac{5k}{4}...\left( iii \right)$
We know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ and ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$.
Using these two identities, we can write
${{\left( a+b \right)}^{2}}-4ab=\left( {{a}^{2}}+{{b}^{2}}+2ab \right)-4ab$
And so, we can easily write
${{\left( a+b \right)}^{2}}-4ab={{a}^{2}}+{{b}^{2}}-2ab$
Thus, we get
${{\left( a+b \right)}^{2}}-4ab={{\left( a-b \right)}^{2}}$
Hence, by using this identity, we can write
${{\left( \alpha +\beta \right)}^{2}}-4\alpha \beta ={{\left( \alpha -\beta \right)}^{2}}$
Hence, by using equation (ii) and equation (iii), we get
${{\left( \alpha -\beta \right)}^{2}}={{\left( \dfrac{5k+1}{4} \right)}^{2}}-4\left( \dfrac{5k}{4} \right)$
Thus, on solving further, we can simplify the above equation as
${{\left( \alpha -\beta \right)}^{2}}=\left( \dfrac{25{{k}^{2}}+1+10k}{16} \right)-5k$
By taking the LCM on the right had side of this equation, we get
${{\left( \alpha -\beta \right)}^{2}}=\dfrac{25{{k}^{2}}+1+10k-80k}{16}$
Thus, we now have the following,
${{\left( \alpha -\beta \right)}^{2}}=\dfrac{25{{k}^{2}}+1-70k}{16}...\left( iv \right)$
In our question, we are given that the roots differ by 1. So, we have $\alpha -\beta =1$.
On squaring both sides, we get
\[{{\left( \alpha -\beta \right)}^{2}}={{\left( 1 \right)}^{2}}\]
Hence, we get
\[{{\left( \alpha -\beta \right)}^{2}}=1...\left( v \right)\]
Now, we can easily equate the right hand side of equation (iv) with that of equation (v),
$\dfrac{25{{k}^{2}}+1-70k}{16}=1$
Thus, we have
$25{{k}^{2}}+1-70k=16$
Or, we can write
$25{{k}^{2}}-70k-15=0$
Let us take a common of 5 from the left hand side,
$5\left( 5{{k}^{2}}-14k-3 \right)=0$
And thus, we have
$5{{k}^{2}}-14k-3=0$
We can easily write it as
$5{{k}^{2}}-15k+k-3=0$
Taking appropriate commons, we get
$5k\left( k-3 \right)+1\left( k-3 \right)=0$
So, we get,
$\left( k-3 \right)\left( 5k+1 \right)=0$
So, we can now write,
$k=3,-\dfrac{1}{5}$

Hence, the values of $k\text{ are }3\text{ and }-\dfrac{1}{5}$.

Note: We can also solve this question by finding the roots of the original given quadratic equation in $x$. We can, then, subtract these two roots and equate the difference to 1. And we can, further, solve our evaluated equation to find the values of $k$.