- compare equations - x-2 - 2 -3-x - 3 - 0 and -160- - ax-2 - bx - c - 0 find - the values of -160- a- b- c -160- - Find value - of b-2 - 4ac - Write formula - to solve - quadratic - -160- equation - Substitute - values of -160- - a- b- c and- -160- find roots -160-Solve it-

Given, x^2+2√(3)x+3=0 a=1,b=2√(3),c=3 b^2 4ac =(2√(3))^2 4(1)(3) =12 12=0 x= b±√(b^2 4ac)2a = 2√(3)±02 x= √(3) ...

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Byju's Answer

Standard XII

Mathematics

Quadratic Formula for Finding Roots

compare equa...

Question

$\begin{matrix} c o m p a r e e q u a t i o n s x^{2} + 2 \sqrt{3} x + 3 = 0 a n d a x^{2} + b x + c = 0 f i n d t h e v a l u e s o f a, b, c \end{matrix} \to \begin{matrix} F i n d v a l u e o f b^{2} - 4 a c \end{matrix} \to \begin{matrix} W r i t e f o r m u l a t o s o l v e q u a d r a t i c e q u a t i o n \end{matrix} \to \begin{matrix} S u b s t i t u t e v a l u e s o f a, b, c a n d f i n d r o o t s \end{matrix}$
Solve it.

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Solution

Given,

$x^{2} + 2 \sqrt{3} x + 3 = 0$

$a = 1, b = 2 \sqrt{3}, c = 3$

$b^{2} - 4 a c$

$= (2 \sqrt{3})^{2} - 4 (1) (3)$

$= 12 - 12 = 0$

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

$= \frac{- 2 \sqrt{3} \pm 0}{2}$

$x = - \sqrt{3}$

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Similar questions

Q. When will the quadratic equation

a x^{2} + b x + c = 0

NOT have Real Roots?

Q. What are the roots of the quadratic equation

a x^{2} + b x + c = 0

Q. If the roots of

a x^{2} + b x + c = 0

are

α, β

and the roots of

A X^{2} + B x + C = 0

are

(α - k), (β - k)

. Then

(\frac{B^{2} - 4 A C}{b^{2} - 4 a c})

is equal to

The roots of a quadratic equation $a x^{2} + b x + c = 0$ are given by $\frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ , provided $b^{2} - 4 a c \geq 0$ .

Q. If

b^{2} - 4 a c > 0

, then the graph of

y = a x^{2} + b x + c

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Given,

$x^{2} + 2 \sqrt{3} x + 3 = 0$

$a = 1, b = 2 \sqrt{3}, c = 3$

$b^{2} - 4 a c$

$= (2 \sqrt{3})^{2} - 4 (1) (3)$

$= 12 - 12 = 0$

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

$= \frac{- 2 \sqrt{3} \pm 0}{2}$

$x = - \sqrt{3}$

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compare equationsx2+2√3x+3=0 andax2+bx+c=0 findthe values of a,b,c→Find valueof b2−4ac→Write formulato solvequadraticequation→Substitutevalues ofa,b,c andfind rootsSolve it.

Given,x2+2√3x+3=0a=1,b=2√3,c=3b2−4ac=(2√3)2−4(1)(3)=12−12=0x=−b±√b2−4ac2a=−2√3±02x=−√3

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EXAM PREPARATION

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FREE TEXTBOOK SOLUTIONS

STATE BOARDS

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Given,

$x^{2} + 2 \sqrt{3} x + 3 = 0$

$a = 1, b = 2 \sqrt{3}, c = 3$

$b^{2} - 4 a c$

$= (2 \sqrt{3})^{2} - 4 (1) (3)$

$= 12 - 12 = 0$

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

$= \frac{- 2 \sqrt{3} \pm 0}{2}$

$x = - \sqrt{3}$