Question

help me with this sum
using quadratic formola​

Answer 1

Step-by-step explanation:

4-11x=3x²

3x²=4-11x

3x²+11x-4=0

3x²+12x-x-4=0

3x(x+4)-(x+4)=0

(3x-1)(x+4)=0

x=1/3 or x=-4

Answer 2

Question :-

Solve this using quadratic formula -

4 - 11x = 3x²

Solution :-

=> 4 - 11x = 3x²

=> -3x² + 4 - 11x = 0

=> -3x² - 11x + 4 = 0

In the given equation,

a = -3,

b = -11,

c = 4

Quadratic Formula -

x = $\dfrac{-b + \sqrt{b^{2}- 4ac } }{2a}$  OR  $\dfrac{-b - \sqrt{b^{2}- 4ac } }{2a}$

Substituting values from the equation in the Quadratic Formula -

So, x = $\dfrac{-(-11) + \sqrt{(-11)^{2}- (4\times -3\times 4) } }{2\times -3}$  

                               OR

           $\dfrac{-(-11) - \sqrt{(-11)^{2}- (4\times -3\times 4) } }{2\times -3}$

=> x = $\dfrac{11 + \sqrt{121- (-48) } }{-6}$  OR  $\dfrac{11 - \sqrt{121- (-48) } }{-6}$

=> x = $\dfrac{11 + \sqrt{121+48 } }{-6}$  OR  $\dfrac{11 - \sqrt{121+48} }{-6}$

=> x = $\dfrac{11 + \sqrt{169 } }{-6}$  OR  $\dfrac{11 - \sqrt{169} }{-6}$

=> x = $\dfrac{11 +13 }{-6}$  OR  $\dfrac{11 -13}{-6}$

=> x = $\dfrac{24}{-6}$  OR  $\dfrac{-2}{-6}$

=> x = $\dfrac{-4}{1}$  OR  $\dfrac{1}{3}$

=> x = -4  OR  $\dfrac{1}{3}$

∴ The roots of given equation are :- -4 and  $\dfrac{1}{3}$.