Question

If roots of the given quadratic $2x^2-9x+7=0$ are $p,q.$ Then the equation whose roots are

• A. $50x^2-45x+7=0$
• B. $2x^2-43x+203=0$
• C. $2x^2-x-3=0$
• D. $2x^2+9x+7=0$

Answer 1

Given: $2x^2-9x+7=0;$ roots are $p,q$
$\text{Sum of roots}=p+q=\frac{9}{2}\cdots \left(i\right)$

$\text{Product of roots}=p.q=\frac{7}{2}\cdots \left(ii\right)$

Now we will solve for the one by one

$a)-p,-q$
$\therefore \text{Sum of roots}=-p-q=\frac{-b}{a}$

$\Rightarrow \text{Sum of roots}=-(p+q)=\frac{-b}{a}$

$\Rightarrow \text{Sum of roots}=\frac{-b}{a}=-(\frac{9}{2})\left[\text{From}\right(i\left)\right]$

$\Rightarrow \text{Sum of roots}=\frac{-b}{a}=-\frac{9}{2}\cdots \left(iii\right)$

Now, $\text{Product of roots}=(-p)(-q)=\frac{c}{a}$

$\Rightarrow \text{Productof roots}=\frac{c}{a}=p.q$

$\Rightarrow \text{Productof roots}=\frac{c}{a}=\frac{7}{2}\cdots \left(iv\right)\left[\text{From}\right(ii\left)\right]$

Now we have the sum , products of roots for the required equation.And we can write the equation as: $x^2-\left(\text{sum of roots}\right)x+\left(\text{product of roots}\right)=0$

$\therefore x^2-(-\frac{9}{2})x+\left(\frac{7}{2}\right)=0$

$\Rightarrow 2x^2+9x+7=0$

$b)p-2,q-2$
$\therefore \text{Sum of roots}=p-2+q-2=\frac{-b}{a}$

$\Rightarrow \text{Sum of roots}=(p+q)-4=\frac{-b}{a}$

$\Rightarrow \text{Sum of roots}=\frac{-b}{a}=(\frac{9}{2})-4\left[\text{From}\right(i\left)\right]$

$\Rightarrow \text{Sum of roots}=\frac{-b}{a}=\frac{1}{2}\cdots \left(v\right)$

Now, $\text{Product of roots}=(p-2)(q-2)=\frac{c}{a}$

$\Rightarrow \text{Productof roots}=\frac{c}{a}=p.q-2(p+q)+4$

$\Rightarrow \text{Productof roots}=\frac{c}{a}=\frac{7}{2}-2\left(\frac{9}{2}\right)+4\left[\text{From}\right(i),(ii\left)\right]$

$\Rightarrow \text{Productof roots}=\frac{c}{a}=-\frac{3}{2}\cdots \left(vi\right)$

Now we have the sum , products of roots for the required equation.And we can write the equation as: $x^2-\left(\text{sum of roots}\right)x+\left(\text{product of roots}\right)=0$

$\therefore x^2-\left(\frac{1}{2}\right)x+(-\frac{3}{2})=0$

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

$c)\frac{p}{5},\frac{q}{5}$
$\therefore \text{Sum of roots}=\frac{p}{5}+\frac{q}{5}=\frac{-b}{a}$

$\Rightarrow \text{Sum of roots}=\frac{p+q}{5}=\frac{-b}{a}$

$\Rightarrow \text{Sum of roots}=\frac{-b}{a}=(\frac{(\frac{9}{2})}{5})\left[\text{From}\right(i\left)\right]$

$\Rightarrow \text{Sum of roots}=\frac{-b}{a}=\frac{9}{10}\cdots \left(vii\right)$

Now, $\text{Product of roots}=\frac{p}{5}.\frac{q}{5}=\frac{c}{a}$

$\Rightarrow \text{Productof roots}=\frac{c}{a}=\frac{p.q}{25}$

$\Rightarrow \text{Productof roots}=\frac{c}{a}=(\frac{(\frac{7}{2})}{25})\left[\text{From}\right(ii\left)\right]$

$\Rightarrow \text{Productof roots}=\frac{c}{a}=\frac{7}{50}\cdots \left(viii\right)$

Now we have the sum , products of roots for the required equation.And we can write the equation as: $x^2-\left(\text{sum of roots}\right)x+\left(\text{product of roots}\right)=0$

$\therefore x^2-\left(\frac{9}{10}\right)x+\left(\frac{7}{50}\right)=0$

$\Rightarrow 50x^2-45x+7=0$

$d)3p+4,3q+4$
$\therefore \text{Sum of roots}=3p+4+3q+4=\frac{-b}{a}$

$\Rightarrow \text{Sum of roots}=3(p+q)+8=\frac{-b}{a}$

$\Rightarrow \text{Sum of roots}=\frac{-b}{a}=3(\frac{9}{2})+8\left[\text{From}\right(i\left)\right]$

$\Rightarrow \text{Sum of roots}=\frac{-b}{a}=\frac{43}{2}\cdots \left(ix\right)$

Now, $\text{Product of roots}=(3p+4)(3q+4)=\frac{c}{a}$

$\Rightarrow \text{Productof roots}=\frac{c}{a}=9p.q+12(p+q)+16$

$\Rightarrow \text{Productof roots}=\frac{c}{a}=9(\frac{7}{2})+12(\frac{9}{2})+16\left[\text{From}\right(ii\left)\right]$

$\Rightarrow \text{Productof roots}=\frac{c}{a}=\frac{203}{2}\cdots \left(x\right)$

Now we have the sum , products of roots for the required equation.And we can write the equation as: $x^2-\left(\text{sum of roots}\right)x+\left(\text{product of roots}\right)=0$

$\therefore x^2-\left(\frac{43}{2}\right)x+\left(\frac{203}{2}\right)=0$

$\Rightarrow 2x^2-43x+203=0$