Solve the equation x, squared, minus, 11, x, plus, 41, equals, 19x

Question

2
−11x+41=19 to the nearest tenth.

Answer 1

To solve the equation $x^2-11x+41=19$, we first need to rearrange it into standard form. We'll subtract 19 from both sides:

$$x^2-11x+41-19=0$$

This simplifies to:

$$x^2-11x+22=0$$

Next, we can use the quadratic formula to solve for $x$. The quadratic formula is given by:

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

Here, $a=1$, $b=-11$, and $c=22$.

Now, we can substitute these values into the formula:

1. Calculate $b^2-4ac$:

$$(-11{)}^2-4(1)(22)=121-88=33$$

2. Substitute into the quadratic formula:

$$x=\frac{-(-11)\pm \sqrt{33}}{2(1)}=\frac{11\pm \sqrt{33}}{2}$$

Next, we find the approximate value of $\sqrt{33}$.

$$\sqrt{33}\approx 5.744$$

Now, substituting this value back into the equation:

$$x=\frac{11\pm 5.744}{2}$$

Calculating the two possible values for $x$:

1. First solution:

$$x_1=\frac{11+5.744}{2}=\frac{16.744}{2}\approx 8.372$$

2. Second solution:

$$x_2=\frac{11-5.744}{2}=\frac{5.256}{2}\approx 2.628$$

Thus, the solutions for $x$ rounded to the nearest tenth are:

$$x\approx 8.4\text{and}x\approx 2.6$$