Question

1. find the real roots for $y = x^2 - 6x$. use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

(6,0)
(6,6)
(0,0)
(0, -6)
(-6,0)

1. solve.

$|x + 3| < 7$
$-10 < x < 4$
$4 < x < 10$
$x > |-10|$
$x < |4|$

Explanation:

For Question 14:

Step1: Identify a, b, c

For $y=x^2-6x$, set $y=0$, so $ax^2+bx+c=0$ gives $a=1$, $b=-6$, $c=0$.

Step2: Substitute into quadratic formula

$$x=\frac{-(-6)\pm\sqrt{(-6)^2-4(1)(0)}}{2(1)}$$

Step3: Simplify the expression

$$x=\frac{6\pm\sqrt{36}}{2}=\frac{6\pm6}{2}$$

Step4: Calculate two roots

First root: $\frac{6+6}{2}=6$, second root: $\frac{6-6}{2}=0$. Roots as points are $(6,0)$ and $(0,0)$.

Step1: Rewrite absolute value inequality

For $|x+3|<7$, apply $|A|$$-7 < x+3 < 7$$

Step2: Isolate x by subtracting 3

$$-7-3 < x < 7-3$$

Step3: Simplify the inequality

$$-10 < x < 4$$

Answer:

(6, 0), (0, 0)

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For Question 15: