First, we need to rewrite the equation in standard form:
$x^2 - 21x + 4x = 0$
Combine like terms:
$x^2 - 17x = 0$
Now we can use the quadratic formula:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
In this case, $a=1$, $b=-17$, and $c=0$. Plugging those values in, we get:
$x = \frac{-(-17) \pm \sqrt{(-17)^2 - 4(1)(0)}}{2(1)}$
Simplify:
$x = \frac{17 \pm \sqrt{289}}{2}$
$x = \frac{17 \pm 17}{2}$
$x_1 = \frac{17+17}{2} = 17$
$x_2 = \frac{17-17}{2} = 0$
Therefore, the solution is: $x=17$ or $x=0$.
Answer: $\boxed{\text{A. }7, 3}$
Use the quadratic formula to solve the equation. If necessary, round to the nearest hundredth.
x squared minus 21 x equals negative 4 x
A. 7, 3
B. negative 7, 3
C. 7, negative 3
D. negative 7, negative 3
3 answers
Which kind of function best models the data in the table? Use differences or ratios.
x y
0 1.7
1 6.8
2 27.2
3 108.8
4 435.2
A. linear
B. quadratic
C. exponential
D. none of the above
21 / 21
x y
0 1.7
1 6.8
2 27.2
3 108.8
4 435.2
A. linear
B. quadratic
C. exponential
D. none of the above
21 / 21
Looking at the ratios of consecutive $y$-values, we see that they are increasing by a factor of 4 each time:
$\frac{6.8}{1.7} = 4$
$\frac{27.2}{6.8} = 4$
$\frac{108.8}{27.2} = 4$
$\frac{435.2}{108.8} = 4$
This indicates that the function is exponential.
To confirm, we can take the ratios of the consecutive differences of $y$-values:
$\frac{6.8-1.7}{1-0} = 5.1$
$\frac{27.2-6.8}{2-1} = 20.4$
$\frac{108.8-27.2}{3-2} = 81.6$
$\frac{435.2-108.8}{4-3} = 326.4$
These ratios are not constant, so the function is not linear or quadratic.
Therefore, the answer is $\boxed{\text{C. exponential}}$.
$\frac{6.8}{1.7} = 4$
$\frac{27.2}{6.8} = 4$
$\frac{108.8}{27.2} = 4$
$\frac{435.2}{108.8} = 4$
This indicates that the function is exponential.
To confirm, we can take the ratios of the consecutive differences of $y$-values:
$\frac{6.8-1.7}{1-0} = 5.1$
$\frac{27.2-6.8}{2-1} = 20.4$
$\frac{108.8-27.2}{3-2} = 81.6$
$\frac{435.2-108.8}{4-3} = 326.4$
These ratios are not constant, so the function is not linear or quadratic.
Therefore, the answer is $\boxed{\text{C. exponential}}$.