Question

ax²+2bx+c=0(a≠0) dighat সমীকরণের বীজ দুটি বাস্তব ও সমান হলে b²=ac প্রমাণ করো।​

Answer 1

❏ Question:-

If two roots of the quadratic equation ax²+2bx+c=0 are equal and real then prove that b²=ac.

❏ Solution:-

For a quadratic equation ax²+bx+c=0 ,

the solution or the roots are given by ,

(fr0m Sridhar Achariya Law/Formula),

$\sf\implies x_1=\dfrac{-b+\sqrt{b^2-4ac}}{2a}$

and ,

$\sf\implies x_2=\dfrac{-b-\sqrt{b^2-4ac}}{2a}$

Now, according to the question the roots are real and equal,

i.e, $\sf\implies x_1=x_2$

$\sf\implies \dfrac{-b+\sqrt{b^2-4ac}}{\cancel{2a}}=\dfrac{-b-\sqrt{b^2-4ac}}{\cancel{2a}}$

$\sf\implies \cancel{-b}+\sqrt{b^2-4ac}=\cancel{-b}-\sqrt{b^2-4ac}$

$\sf\implies \sqrt{b^2-4ac}=-\sqrt{b^2-4ac}$

$\sf\implies\sqrt{b^2-4ac}+\sqrt{b^2-4ac}=0$

$\sf\implies 2\sqrt{b^2-4ac}=0$

$\sf\implies \sqrt{b^2-4ac}=0$

$\sf\implies b^2-4ac=0$

$\sf\implies\boxed{\red{\large{ b^2=4ac}}}$

Here the given equation is ,

ax²+2bx+c=0

so , from the above Condition,

$\sf\implies{ (2b)}^2-4ac=0$

$\sf\implies 4b^2=4ac$

$\sf\implies b^2=ac$

$\sf\implies\boxed{\red{\large{ b^2=ac}}}$ (proved)

Answer 2

Question :--- if Roots of Equation ax² + 2bx +c = 0 (where a ≠0) are Equal , than prove that, b² = ac ? .

Solution :--

Since , roots of the Equation are Given Equal, we can assume that, let both the roots are y.

So, we can say that, now,

→ (x-y)(x-y) = 0

→ x² -2xy + y² = 0

Now, given Equation is = ax² + 2bx + c = 0

Comparing both now, we get,

→ x² -2xy + y² = 0 = ax² + 2bx + c

→ a = 1

→ b = (-y)

→ c = y²

So, Putting all these value in b² = ac now, we get,

→ b² = a * c

→ (-y)² = 1 * y²

→ y² = y² [✪✪ Hence Proved ✪✪]

Hence, we can say that, if Roots of Equation ax² + 2bx +c = 0 (where a ≠0) are Equal , than b² is Equal to ac .