Mathematics - Quadratic Equation Question with Solution | TestHub

Let ${\mathrm{x}}^2+19\mathrm{x}+89={\lambda }^2$

$\Rightarrow {\mathrm{x}}^2+19\mathrm{x}+\left(89-{\lambda }^2\right)=0$ should have integral roots

∴ D should be a perfect square.

$\Rightarrow (19{)}^2-4\left(89-{\lambda }^2\right)=$ Perfect square

$\Rightarrow (19{)}^2-4\left(89-{\lambda }^2\right)=$ Perfect square $\Rightarrow \left({\mathrm{m}}^2-4{\lambda }^2\right)=5\Rightarrow (\mathrm{m}-2\lambda )(\mathrm{m}+2\lambda )=5$

$\therefore (\mathrm{m}-2\lambda =5,\mathrm{m}+2\lambda =1)$

or $(m-2\lambda =-5,m+2\lambda =-1)$

$\Rightarrow (\mathrm{m}-2\lambda =-5,\mathrm{m}+2\lambda =-1)$

$\Rightarrow \mathrm{m}=3,-3,\lambda =1,-1$

For $\lambda =\pm 1$ equation becomes ${\mathrm{x}}^2+19\mathrm{x}+88=0$

$\Rightarrow (\mathrm{x}+11)(\mathrm{x}+8)=0$

$\Rightarrow \mathrm{x}=-8,-11$.

Thus, required values of $x$ are $-8,-11$.