Question

roots of the equation (a+b-c)x²-2ax+(a-b+c)=0 , (a,b,c belongs to= Q)​

Answer 1

Step-by-step explanation:

3qx²)2qx+(3q)=0

then use law of equation in algebraic equations .then

ans=

Answer 2

SOLUTION

It is clear that the given equation is a quadratic equation

Given expression,

$(a + b - c) {x}^{2} - 2ax + (a - b + c) = 0$

In this given expression,

a , b and c are Real and rational parts

Therefore, a, b and c are constants

Thus,

This expression is of the form,

$a {x}^{2} + bx + c = 0$

Where, a=(a+b-c)

b=-2a

c=(a-b+c)

Let us factorize the given expression using Completing square method,

$\frac{ {ax}^{2} }{a} + \frac{bx}{a} + \frac{c}{a} = \frac{0}{a} \\ = > {x}^{2} + \frac{b}{a} x = - \frac{c}{a} \\ = > {x}^{2} + 2 \times \frac{b}{2a} \times x + (\frac{b}{2a})^{2} = {( \frac{b}{2a}) }^{2} - \frac{c}{a} \\ = > {(x + \frac{b}{2a}) }^{2} = \frac{ {b}^{2} }{4 {a}^{2} } - \frac{c}{a} \\ = > {(x + \frac{b}{2a} )}^{2} = \frac{ {b}^{2} - 4ac}{4 {a}^{2} } \\ = > x + \frac{b}{2a} = \frac{ \frac{ + }{ } \sqrt{ {b}^{2} - 4ac} }{2a} \\ = > x = \frac{ - b + \sqrt{ {b}^{2} - 4ac} }{2a} \: or \: x = \frac{ - b - \sqrt{ {b}^{2} - 4ac} }{2a}$

Now , Just Substitute the values of a, b and c in the value of x

Hence, We get the required roots of the given expression as

$\frac{ - ( - 2a) + \sqrt{ {( - 2a)}^{2} - 4(a + b - c)(a - b + c) } }{2a} \: and \: \frac{ - ( - 2a) - \sqrt{( { - 2a)}^{2} - 4(a + b - c)(a - b + c)} }{2a} \\ \frac{ 2a + 2 \sqrt{{</strong><strong>(</strong><strong>{b} - {c}</strong><strong>)</strong><strong>^{2} }} }{2a} \: and \: \frac{2a - 2 \sqrt{ </strong><strong>(</strong><strong>{b} - {c}</strong><strong>)</strong><strong>^{2} } }{2a} \\ </strong><strong>\</strong><strong>frac</strong><strong>{ { </strong><strong>a</strong><strong>+</strong><strong>{b}- {c} } }{a} \: and \: \frac{ {</strong><strong>a-</strong><strong> {b} </strong><strong>+</strong><strong>{c} } }{a}$

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