Question

if x^3-5x^2+2x=0 then find the value of x ​

Answer 1

Answer:

Solve the above equation as follows:

x

3

−5x

2

+2x−10=0

x

2

(x−5)+2(x−5)=0

x

2

+2=0 and x−5=0

x

2

=−2 and x=5

Since x

2

=−2 implies the imaginary value of x, therefore real value of x is 5.

Answer 2

Question :-

$\mathsf{if \: {x}^{3} - 5 \: {x}^{2} \: + \: 2 \: then \:find \: the }$

$\mathsf{value \: of \: x \: if \: x \: = \: 0 }$

To Find :-

$\mathsf{value \: of \: x }$

Given :-

$\boxed{\mathsf{ x \: = \: 0}}$

Solution :-

Put the value of x = 0 in the equation .

$\mathsf{:\longrightarrow \: {x}^{3} - 5 \: {x}^{2} \: + \: 2 }$

$\mathsf{:\longrightarrow \: ({0})^{3} - 5 \: \times \:( {0})^{2} \: + \: 2 }$

$\mathsf{:\longrightarrow \: 0- 5 \: \times \: 0\: + \: 2 }$

$\mathsf{:\longrightarrow \: 5 \: \times \: 0\: + \: 2 }$

$\mathsf{:\longrightarrow \:\: 0\: + \: 2 }$

$\mathsf{:\longrightarrow \: 2 }$

$\mathsf{so \: the \: value \: of \: x \: = \: 2}$

Final Answer:-

$\mathsf{so \: the \: value \: of}$

$\boxed{\mathsf{ \: x \: = \: 2}}$