Question

If the roots of the quadratic equation ax2 - 6x + 3 = 0 are equal then find
the value of 2a
(i)3
(ii)6
(iii)2
(iv)4​

Answer 1

$\underline{\textsf{Given:}}$

$\textsf{Roots of the equatiion}$

$\mathsf{ax^2-6x+3=0\;are\;equal}$

$\underline{\textsf{To find:}}$

$\textsf{The value of 2a}$

$\underline{\textsf{Solution:}}$

$\mathsf{Consider,\;ax^2-6x+3=0}$

$\mathsf{Here\;a='a',\;b=-6,\;c=3}$

$\mathsf{Since\;the\;roots\;of\;ax^2-6x+3=0\;are\;equal,we\;have}$

$\mathsf{b^2-4ac=0}$

$\mathsf{(-6)^2-4{\times}a{\times}3=0}$

$\mathsf{36-12a=0}$

$\mathsf{-12a=-36}$

$\mathsf{a=\dfrac{-36}{-12}}$

$\implies\boxed{\mathsf{a=3}}$

$\underline{\textsf{Answer:}}$

$\mathsf{The\;value\;of\;2a\;is\;6}$

$\underrline{\textsf{Find more:}}$

If the roots of x^2 -bx/ax-c = k-1/k+1 are numerically equal and opposite in sign then k=?​

https://brainly.in/question/16638285

If x^2-bx/ax-c =m-1/m+1

has roots which are numerically equal but of opposite sign, then the value of'm'must be

(A) a-b/a+b

(B) a+b/a-b

(C) O

(D) 1

soneone please answer this

https://brainly.in/question/9525569

Answer 2

HELLO DEAR,

GIVEN:- A quadratic equation

ax² - 6x + 3 = 0

And their roots are equal.

To find the value of 2a.

SOLUTION:- ax² - 6x + 3 = 0

We know that if D < 0 , the roots are imaginary.

if D > 0 , the roots are real.

if D = 0 , the roots are equal

Where D = b² - 4ac

So, in quadratic equation ax²- 6x + 3 = 0

a= a , b= -6 , c= 3

By using property D = 0 if roots are equal,

⇒ D= 0

⇒ b² - 4ac = 0

⇒ (-6)² - 4a(3) = 0

⇒ 36 - 12a = 0

⇒ -12a = -36

⇒ a= (-36)/(-12)

⇒ a= 3

So, the value of 2a = 2 × 3

2a = 6.

Therefore ii) 6 is correct option.

I HOPE IT'S HELP YOU DEAR,

THANKS.