Question 13/09

Write the equation as: $\sqrt{\color{#20A900}{3x^2-8x+1}}+\sqrt{3(\color{#20A900}{x^2-8x+1})-11}=3$ Substitute $\color{#20A900}{3x^2-8x+1=t}$ s.t eqn is:

$\large\sqrt{\color{#20A900}{t}}+\sqrt{3\color{#20A900}{t}-11}=3..... (1)$ Subsracting $\sqrt{ \color{#20A900}{t}}$ from both sides and squaring: $\large\implies 3t-11=9+t-6\sqrt{t}$ $\large\implies (\color{#20A900}{t}-10)^2=(-3\sqrt{\color{#20A900}{t}})^2$ $\large\implies \color{#20A900}{t}^2-29\color{#20A900}{t}+100=0$ $\large\implies \color{#20A900}{t}=4,25$

However we can reject $\color{#20A900}{t}=25$ since it doesn't satisfy original eqn $(1)$ .

Thus, $\large\color{#20A900}{t}=4=\color{#20A900}{3x^2-8x+1}$ $\large\implies 3x^2-8x-3=0$ $\large\implies x=3,-1/3$ And since $3=\large{\color{#0C6AC7}{-9}}\times\frac{-1}3$ hence answer is $\large\color{#0C6AC7}{\boxed{-9}}$ .

$\begin{aligned} \sqrt{3x^2-8x+1} + \sqrt{9x^2-24x-8} & = 3 \\ \sqrt{\color{#3D99F6}{3x^2-8x+1}} + \sqrt{3(\color{#3D99F6}{3x^2-8x+1})-11} & = 3 \quad \quad \small \color{#3D99F6}{\text{Let }y^2 = 3x^2-8x+1} \\ \implies \sqrt{\color{#3D99F6}{y^2}} + \sqrt{3\color{#3D99F6}{y^2}-11} & = 3 \\ \sqrt{3y^2-11} & = 3 - y \\ 3y^2 - 11 & = y^2 - 6y + 9 \\ 2y^2 + 6 y - 20 & = 0 \\ y^2 + 3 y - 10 & = 0 \\ (y+5)(y-2) & = 0 \\ \implies y & = \begin{cases} \color{#D61F06}{-5} & \color{#D61F06}{\implies \sqrt{(-5)^2} + \sqrt{3(-5)^2-11} \ne 3} & \color{#D61F06}{\text{rejected}} \\ \color{#3D99F6}{2} & \color{#3D99F6}{\implies \sqrt{(2)^2} + \sqrt{3(2)^2-11} = 3} & \color{#3D99F6}{\text{accepted}} \end{cases} \end{aligned}$

$\begin{aligned} \implies 3x^2-8x+1 & = y^2 = 4 \\ 3x^2-8x-3 & = 0 \\ (3x+1)(x-3) & = 0 \\ \implies x & = \begin{cases} -\frac{1}{3} \\ 3 \end{cases} \\ \implies k & = \frac{3}{-\frac{1}{3}} = \boxed{- 9} \end{aligned}$