1 Solve for x 1.1.1 (2x−1)(x−1)=0 1.1.2 2x2−3x−1=0 (Correct to TWO decimal places) 1.1.3 x(2x+1)-3<=0 1.2 Solve for x and y

The solutions are $x=\frac{1}{2}$ or $x=1$.

1.1.2

To solve $2x^2-3x-1=0$, we use the quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$.

For $2x^2-3x-1=0$, we have $a=2$, $b=-3$, $c=-1$.

Rounding to two decimal places, we get $x \approx 1.78$ or $x \approx -0.28$.

1.1.3

To solve the inequality $x(2x+1)-3 \le 0$, we first expand and rearrange it into standard quadratic form.

We can factor the quadratic expression:

Setting each factor to zero gives the roots:

The roots are $x=-\frac{3}{2}$ and $x=1$.

Since the coefficient of $x^2$ is positive ($2 > 0$), the parabola opens upwards. The inequality $2x^2+x-3 \le 0$ means we are looking for the values of $x$ where the parabola is below or on the x-axis. This occurs between and including the roots.
Therefore, the solution is $-\frac{3}{2} \le x \le 1$.