# if the roots of the Q.E. are equal then prove that b²=4ac.10 class quadratic equations.

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ax^2+bx+c = has equal roots.

We know that if roots are equal say x1 and x1 then we can write the equation as:

(x-x1)(x-x1) = 0 and ax^2+bx+c = 0 same equations.

So a(x-x1)(x-x1) = ax^2+bx+c must be an identity.

Expanding the left, we get:

ax^2 - 2ax1*x+ax1^2 = ax^2+bx+c. is an identity.

So we can equate the like terms on both sides.

-2ax1 = b ..(3)

ax1^2 = c....(4)

We eliminate x1 between (3) and (4).

(-2ax1)^2/ax1 = b^2/c

4a^2/a = b^2/c

4ac= b^2 . Or b^2-4ac = 0.

The given quadratic equation is ax^2 + bx + c = 0.

The roots of a quadratic equation are given as

R1 = -b + sqrt [ b^2 - 4ac ]

R2 = -b - sqrt [ b^2 - 4ac]

Now it is given that the two roots are equal or R1 = R2

=> -b + sqrt [ b^2 - 4ac ] = -b - sqrt [ b^2 - 4ac]

This is possible only if sqrt [ b^2 - 4ac] = 0 or b^2 - 4ac =0 or b^2 = 4ac.

If the roots of the equation are equal:

x1 = x2,

that means that the discriminant of the quadratic equation is equal to zero.

delta = 0

delta = b^2 - 4ac

b^2 - 4ac = 0

We'll add 4ac both sides:

b^2 = 4ac

Let's see how to find delta. We'll write again the q.e.:

ax^2 + bx + c = 0

We'll factorize:

a(x^2 + bx/a + c/a) = 0

a(x^2 + 2bx/2a + b^2/4a^2 - b^2/4a^2+ c/a) = 0

We notice that we've modified the ratio:

bx/a = 2bx/2a

We've also added and subtracted the quantity b^2/4a^2.

We've completed the square x^2 + 2bx/2a + b^2/4a^2.

x^2 + 2bx/2a + b^2/4a^2 = (x + b/2a)^2

a[(x + b/2a)^2 - (b^2/4a^2 - c/a)] = 0

b^2/4a^2 - c/a = (b^2 - 4ac)/4a^2

b^2 - 4ac = delta

a[(x + b/2a)^2 - (delta)/4a^2] = 0

(x + b/2a)^2 - (delta)/4a^2 = 0

(x + b/2a)^2 = (delta)/4a^2

x + b/2a = sqrt delta/2a

x1 = (-b+sqrt delta)/2a

x2 = (-b-sqrt delta)/2a

When x1 = x2:

(-b+sqrt delta)/2a = (-b-sqrt delta)/2a

-b+sqrt delta = -b-sqrt delta

We'll eliminate like terms:

2sqrt delta = 0

delta = 0

b^2 - 4ac = 0

**b^2 = 4ac**