Question

solve that x^2-5x+10=5(x^2-5x+4)^1/2​

Answer 1

There are four solutions to this equation.

Given that,

$\sf {x}^{2} - 5x + 10 = 5( \sqrt{ {x}^{2} - 5x + 4 } )$

• Putting x² - 5x = y,

$\sf \longrightarrow \: y + 10 = 5 \sqrt{y + 4}$

Squaring both sides,

$\sf \longrightarrow {y}^{2} + 100 + 20y = 25(y + 4)$

$\sf \longrightarrow {y}^{2} + 100 + 20y = 25y + 100$

$\sf \longrightarrow y {}^{2} = 5y$

$\sf \longrightarrow y {}^{2} - 5y = 0$

$\sf \longrightarrow y(y - 5) = 0$

Hence either y = 0, or y = 5

If y = 0,

$\sf \: y = 0$

$\sf \longrightarrow {x}^{2} - 5x = 0$

$\sf \longrightarrow x(x - 5) = 0$

$\sf \longrightarrow \underline{ \underline{ x_{1} = 0}}$

Or,

$\sf \longrightarrow \underline{ \underline{ x_{2} = 5}}$

And if y = 5,

$\sf \: y = 5$

$\sf \longrightarrow {x}^{2} - 5x = 5$

$\sf \longrightarrow {x}^{2} - 5x - 5 = 0$

On comparing with ax² + bx + c = 0,

• a = 1
• b = -5
• c = -5

Using quadratic formula,

$\sf \: x_{3,4} = \dfrac{ - b \pm \sqrt{ {x}^{2} - 4ac } }{2a}$

$\sf \longrightarrow \: x_{3,4} = \dfrac{ - ( - 5) \pm \sqrt{ {( - 5)}^{2} - (4 \times 1 \times - 5) } }{2 \times 1}$

$\sf \longrightarrow \: x_{3,4} = \dfrac{ 5 \pm \sqrt{ {25 + 20} } }{2}$

$\sf \longrightarrow \underline{ \underline{ x_{3} = \dfrac{5 + \sqrt{45} }{2} }}$

$\sf \longrightarrow \underline{ \underline{ x_{4} = \dfrac{5 - \sqrt{45} }{2} }}$