Question

find the roots of 4x^2 + 4root 3x+3 by quadratic formula ​

Answer 1

GIVEN :-

• A quadratic equation 4x² + 4√3 x + 3 = 0.

TO FIND :-

• The roots of the quadratic equation.

SOLUTION :-

$\\ : \implies \displaystyle \sf \:4x ^{2} + 4 \sqrt{3} x + 3 = 0$

Now by using the quadratic formula,

$\\ \\ : \implies \displaystyle \sf \:x = \frac{ - b \pm \sqrt{b ^{2} - 4ac } }{2a}$

• a = 4.
• b = 4√3.
• c = 3.

$\\ \\ : \implies \displaystyle \sf \:x = \frac{ - 4 \sqrt{3} \pm \sqrt{(4 \sqrt{3}) ^{2} - 4 \times 4 \times 3} }{2 \times 4}$

$: \implies \displaystyle \sf \:x = \frac{ - 4 \sqrt{3} \pm \sqrt{16 \sqrt{9} - 48 } }{8}$

$: \implies \displaystyle \sf \:x = \frac{ - 4 \sqrt{3} \pm \sqrt{16 \times 3 - 48} }{8}$

$: \implies \displaystyle \sf \:x = \frac{ - 4 \sqrt{3} \pm \sqrt{0} }{8}$

$: \implies \displaystyle \sf \:x = \frac{ - 4 \sqrt{3} + \sqrt{0} }{8} \: , \: x = \frac{ - 4 \sqrt{3} - \sqrt{0} }{8}$

$: \implies \displaystyle \sf \:x = \frac{ - 4 \sqrt{3} + 0 }{8} \: , \: x = \frac{ - 4 \sqrt{3} - 0 }{8}$

$: \implies \displaystyle \sf \:x = \frac{ - 4 \sqrt{3} }{8} \: , \: x = \frac{ - 4 \sqrt{3} }{8}$

$: \implies \underline{ \boxed{ \displaystyle \sf \:x = \frac{ - \sqrt{3} }{2} \: , \: x = \frac{ - \sqrt{3} }{2} }}$

Answer 2

Answer:

$\huge \bf \: solution$

4x² + 4√3x + 3 = 0

Now

Applying Quardtic formula

x = -b ± √b² - 4ac / 2a

• A = 4
• B = 4√3
• C = 3

x = -4 √3 ± √(4√3)² - 4× 4 × 3 / 2 × 4

x = -4√3 ± √16√9 - 48 /8

x = -4 √3 ± √16 × 3 - 48 /8

x = -4√3 ±0/8

Now,

x = -4 √3 + √0/8 , x = -4 √3 - √0/8

x = -4√3 + 0/8 , x = -4√3 - 0/8

x = -4√3 - 0 /8, x = -4√3-0/8

$\huge \bf \: x \: = \frac{ -\sqrt{3} }{2}, x \: = \frac{ -\sqrt{3} }{2}$