Question

Complete the following table:

1. Character : ____________________

2. Setting : _______________________

3. Problem : _______________________

4. Solution: _________________________​

Answer 1

Explanation:

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Given :-

The quadratic equation (a-b)x²+(b-c)x+(c-a)=0 are equal real roots.

To prove :-

2a=b+c

Theory :-

For a Quadratic equation of the form

ax²+bx+c= 0 , the expression b²-4ac is called the discriminant.

Nature of roots:-

The roots of a quadratic equation can be of three types.

If D>0, the equation has two distinct real roots.

If D=0, the equation has two equal real roots.

If D<0, the equation has no real roots.

Solution :

We have , (a-b)x²+(b-c)x+(c-a)=0

On comparing with the standard form of Quadratic equation ax²+bx+c= 0.

Here ,

a= (a-b)

b= (b-c)

and c = (c-a)

When equation have equal ro ots then,

Discriminant = 0

\sf{\implies ({b-c}) ^{2} - 4(a-b)(c-a)=0 }⟹(b−c)

2

−4(a−b)(c−a)=0

\sf{\implies {b}^{2} + {c}^{2} - 2cb - 4ac + 4 {a}^{2} + 4bc - 4ab = 0 }⟹b

2

+c

2

−2cb−4ac+4a

2

+4bc−4ab=0

\sf{\implies {b}^{2} + {c}^{2} + 4 {a}^{2} + 4bc - 4ac - 4 ab = 0 }⟹b

2

+c

2

+4a

2

+4bc−4ac−4ab=0

\sf{\implies {b}^{2} + {c}^{2} +( {-2a})^{2}+2bc + 2c(-2a) + 2(-2a)b = 0}⟹b

2

+c

2

+(

=0

\sf{\implies b + c -2a = 0 }⟹b+c−2a=0

Hence proved ✔

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Answer 2

Answer:

1 : good character

2: goal setting

3: face the problem

4: thinking is the solution

Explanation:

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