2 distinct real roots

Find " m " such that

(x^{2}-5x+4)^{2}+(2m+1)(x^{2}-5x+4)+3=0 has two distinct Real roots.

4 Answers

36
rahul ·

let t = x2 - 5x + 4

so, t2 + (2m + 1)t + 3 = 0

for this two have two distinct real roots...

D > 0 => (2m + 1)2 - 12 > 0

=> 4m2 + 1 + 4m - 12 > 0

=> 5m2 + 4m - 11 > 0

Now m....

but i think it is incorrect.. :D Just tried my hand

71
Vivek @ Born this Way ·

but i think it is incorrect..

Lagta toh hai!

1708
man111 singh ·

\hspace{-16}$Here $\mathbf{(x^2-5x+4)^2+(2m+1).(x^2-5x+4)+3=0}$\\\\ Now Let $\mathbf{{x^2-5x+4}=t}\;$, Then equation Convert into \\\\ $\mathbf{t^2+(2m+1).t+3=0}$\\\\ So here $\mathbf{D=(2m+1)^2-4.3=4m^2+4m-11}$\\\\ $\boxed{\mathbf{D=4m^2+4m-11}}$\\\\ So $\mathbf{t=\frac{-(2m+1)\pm \sqrt{D}}{2}}$\\\\ $\mathbf{x^2-5x+4=\frac{-(2m+1)\pm \sqrt{D}}{2}}$\\\\ Now Taking $\mathbf{(+ve)}$ sign and $\mathbf{(-ve)}$ sign seperately \\\\ $\mathbf{x^2-5x+4=\frac{-(2m+1)+\sqrt{D}}{2}}$ and $\mathbf{x^2-5x+4=\frac{-(2m+1)-\sqrt{D}}{2}}$\\\\ \textbf{\underline{\underline{For First::::}}}\\\\ $\mathbf{2x^2-10x+8=-(2m+1)+\sqrt{D}\Leftrightarrow 2x^2-10x+\left(9+2m-\sqrt{D}\right)=0}$\\\\ So For Distinct real Roots either Discriminant is $\mathbf{> 0}$ OR $\mathbf{<0}$\\\\ Let $\mathbf{D_{1}=100-4.2.\left(9+2m-\sqrt{D}\right)=100-72-16m+8\sqrt{D}}$\\\\ $\boxed{\mathbf{D_{1}=28-16m+8\sqrt{D}=4.(7-4m+2\sqrt{D})}}$\\\\ Now If $\mathbf{D_{1}>0\Leftrightarrow 7-4m+2\sqrt{D}>0}$\\\\

\hspace{-16}$So $\mathbf{2\sqrt{D}>-\left(7-4m\right)}$\\\\ $\mathbf{2\sqrt{4m^2+4m-11}>-(7-4m)}$\\\\ $\bullet$ If $\mathbf{(7-4m)\geq 0\Leftrightarrow m\geq \frac{7}{4}}$\\\\ So Given Inequality is Valid for $\mathbf{\boxed{\bold{m\geq \frac{7}{4}}}...............(1)}$ \\\\ $\bullet$ If $\mathbf{(7-4m)< 0\Leftrightarrow m< \frac{7}{4}}$\\\\\\ So $\mathbf{2\sqrt{4m^2+4m-11}>-(7-4m)\Leftrightarrow 4.(4m^2+4m-11)<(49+16m^2-56m)}$\\\\ So $\mathbf{m> \frac{93}{72}}$\\\\ So $\mathbf{\boxed{\frac{93}{72}<\bold{m< \frac{7}{4}}}..............(2)}$\\\\ So from equation $\mathbf{(1)}$ and $\mathbf{(2)}$ $\mathbf{\boxed{\bold{m> \frac{93}{74}}}}$ for $\mathbf{D_{1}>0}$\\\\ Now You can calculate for $\mathbf{D_{1}<0}$

\hspace{-16}$\textbf{\underline{\underline{For Second::::}}}\\\\ $\mathbf{2x^2-10x+8=-(2m+1)-\sqrt{D}\Leftrightarrow 2x^2-10x+\left(9+2m+\sqrt{D}\right)=0}$\\\\ So For Distinct real Roots either Discriminant is $\mathbf{> 0}$ OR $\mathbf{<0}$\\\\ Let $\mathbf{D_{2}=100-4.2.\left(9+2m+\sqrt{D}\right)=100-72-16m-8\sqrt{D}}$\\\\ $\boxed{\mathbf{D_{2}=28-16m-8\sqrt{D}=4.(7-4m-2\sqrt{D})}}$\\\\ Now If $\mathbf{D_{2}>0\Leftrightarrow 7-4m-2\sqrt{D}>0}$\\\\ $\mathbf{(7-4m)>2\sqrt{D}\Leftrightarrow 2\sqrt{D}<(7-4m)}$\\\\ $\bullet$ If $\mathbf{(7-4m)\leq 0 }$ and $\mathbf{D=0}$\\\\ No value of $\mathbf{m}$\\\\ $\bullet$ If $\mathbf{(7-4m)> 0\Leftrightarrow \mathbf{m<\frac{7}{4}} }$\\\\ $\mathbf{2\sqrt{D}<(7-4m)}$\\\\ So $\mathbf{m< \frac{93}{72}}$\\\\ So Second Inequality is valid for $\mathbf{\boxed{\bold{m <\frac{93}{74}}}}$\\\\ So $\mathbf{\boxed{\bold{m <\frac{93}{74}}}}$ for $\mathbf{D_{2}>0}$\\\\\\ Now You can calculate for $\mathbf{D_{2}<0}$\\\\\\

\hspace{-16}$Now Given equation has $\mathbf{2}$ distinct Real Roots , Then \\\\ $\bullet$ $\mathbf{D_{1}>0}$ and $\mathbf{D_{2}<0}$\\\\ $\bullet$ $\mathbf{D_{1}<0}$ and $\mathbf{D_{2}>0}$\\\\

agar galti ho to dekh lena yaar....

thanks

71
Vivek @ Born this Way ·

Thanks! Itni mehnat ke baad galti ki jegah kaha!!

But is there any other method available for doing this sum? HS sir Kindly let us know!

Your Answer

Close [X]