No no need to take out the 2
You can factor this by using the ac method

2x^2 + 3x - 5 = 0

multiplying the first and last coefficients 2 * -5 = -10
Now we want to get 2 factors of -10 which will give us +3 when added:-
+5 and -2 seem ok so we write:-

2x^2 - 2x + 5x - 5 = 0

2x(x - 1) + 5 (x - 1) =0

(2x + 5)(x - 1) = 0

Solution is x = -5/2 and x = 1

3 0

4 years ago

What are solutions to the equation sinx cosx = square root 3 over4

andrey2020 [161]

$\sin(x)\cos(x)=\dfrac{\sqrt{3}}{4}\\\\2\sin(x)\cos(x)=\dfrac{\sqrt{3}}{2}\\\\\sin(2x)=\dfrac{\sqrt{3}}{2}\\\\2x=\arcsin\left( \dfrac{\sqrt{3}}{2} \right)\\\\2x=\left\{ \dfrac{\pi}{3},\dfrac{2\pi}{3}\right\}+2n\pi\\\\\boxed{x=\left\{ \dfrac{\pi}{6},\dfrac{\pi}{3}\right\}+n\pi}$

7 0

4 years ago