Question

If x^a - 2ax + a^2 = 0 FIND VALUE OF x/a

Answer 1

Answer:

Hi there !!!

Given,

x - a is a factor of

{x}^{3} - ax {}^{2} + 2x + a - 1

we know,

x - a = 0

x = 0+a = a

Substitung the value of x as a,

we have,

(a)^3 - a(a)^2 + 2(a) + a - 1 =0

a^3 - a^3 + 2a + a - 1=0

3a - 1=0

3a =1

a = 1/3

Thus

a = 1/3

Step-by-step explanation:

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Answer 2

$\huge\mathfrak\blue{Answer:}$

Given:

• We have been given a Quadratic Polynomial
• x² - 2ax + a² = 0

To Find:

• We have to find the value of x/a

Solution:

$\bigstar \: \: \underline{\large\mathfrak\orange{We \: have \: been \: given:}}$

$\hookrightarrow \sf{x^2 - 2ax + a^2 = 0}$

Using $\boxed{\sf{\green{ {( a + b )}^2 = a^2 + b^2 + 2ab }}}$

$\hookrightarrow \sf{ {( x - a )}^2 = 0}$

Taking square root on both sides

$\hookrightarrow \sf{ x - a = \sqrt{0}}$

$\hookrightarrow \sf{ x - a = 0}$

$\hookrightarrow \sf{x = a }$

Dividing Both Sides by a

$\hookrightarrow \sf{\dfrac{x}{a} = 1 }$

:$\implies \boxed{\sf{\red{\dfrac{x}{a} = 1}}}$

Hence value of x/a is 1

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$\boxed{\sf{\pink{Some \: Useful \: Indentities }}}$

$\implies \sf{{(a+b}^{2} = a^2 + b^2 + 2ab}$

$\implies \sf{{(a-b)}^{2} = a^2 + b^2 - 2ab}$

$\implies \sf{{(a+b)}^{3} = a^3 + b^3 + 3ab(a+b)}$

$\implies \sf{{(a-b)}^{3} = a^3 - b^3 - 3ab(a-b)}$

$\implies \sf{a^3 + b^3 = (a+b)( a^2 + b^2 - ab) }$

$\implies \sf{a^3 - b^3= (a-b)(a^2 + b^2 +ab) }$

$\implies \sf{{(a+b+c)}^{2} = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca}$

$\implies \sf{a^3+b^3+c^3 - 3abc= (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca) }$

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