12 multiple-choice questions, progressively harder.
Solve x2−x−20≥0x^2 - x - 20 \ge 0x2−x−20≥0.
Solution
Correct answer: A
Factor to find the roots.
x2−x−20=(x−5)(x+4)=0⇒x=−4 or x=5x^2 - x - 20 = (x - 5)(x + 4) = 0 \Rightarrow x = -4 \text{ or } x = 5x2−x−20=(x−5)(x+4)=0⇒x=−4 or x=5
The upward parabola is positive outside the roots, and the inclusive ≥0\ge 0≥0 keeps them, so x≤−4x \le -4x≤−4 or x≥5x \ge 5x≥5.
Solve x2−3x−10<0x^2 - 3x - 10 < 0x2−3x−10<0.
Correct answer: C
x2−3x−10=(x−5)(x+2)=0⇒x=−2 or x=5x^2 - 3x - 10 = (x - 5)(x + 2) = 0 \Rightarrow x = -2 \text{ or } x = 5x2−3x−10=(x−5)(x+2)=0⇒x=−2 or x=5
The upward parabola is negative between the roots, and the strict symbol excludes them, so −2<x<5-2 < x < 5−2<x<5.
Solve x2−8x+16>0x^2 - 8x + 16 > 0x2−8x+16>0.
Correct answer: B
The quadratic is a perfect square.
x2−8x+16=(x−4)2x^2 - 8x + 16 = (x - 4)^2x2−8x+16=(x−4)2
A square is positive everywhere except where it is zero, so (x−4)2>0(x - 4)^2 > 0(x−4)2>0 holds for every xxx except x=4x = 4x=4, giving x<4x < 4x<4 or x>4x > 4x>4.
Solve x2−10x+25≤0x^2 - 10x + 25 \le 0x2−10x+25≤0.
x2−10x+25=(x−5)2x^2 - 10x + 25 = (x - 5)^2x2−10x+25=(x−5)2
A square is never negative, so (x−5)2≤0(x - 5)^2 \le 0(x−5)2≤0 can only hold where it equals 000, at the single point x=5x = 5x=5.
Solve 2x2−x−6≤02x^2 - x - 6 \le 02x2−x−6≤0.
Correct answer: D
Factor the quadratic with a leading coefficient of 222.
2x2−x−6=(2x+3)(x−2)=0⇒x=−32 or x=22x^2 - x - 6 = (2x + 3)(x - 2) = 0 \Rightarrow x = -\tfrac{3}{2} \text{ or } x = 22x2−x−6=(2x+3)(x−2)=0⇒x=−23 or x=2
The upward parabola is negative between the roots, and the inclusive symbol keeps the endpoints, so −32≤x≤2-\tfrac{3}{2} \le x \le 2−23≤x≤2.
Solve x2−3<0x^2 - 3 < 0x2−3<0.
The roots come from setting the quadratic to 000.
x2−3=0⇒x2=3⇒x=±3x^2 - 3 = 0 \Rightarrow x^2 = 3 \Rightarrow x = \pm\sqrt{3}x2−3=0⇒x2=3⇒x=±3
The upward parabola is negative between the roots, so −3<x<3-\sqrt{3} < x < \sqrt{3}−3<x<3.
Solve x2−6x+4≥0x^2 - 6x + 4 \ge 0x2−6x+4≥0.
Use the quadratic formula with a=1a = 1a=1, b=−6b = -6b=−6, c=4c = 4c=4.
x=6±36−162=3±5x = \frac{6 \pm \sqrt{36 - 16}}{2} = 3 \pm \sqrt{5}x=26±36−16=3±5
The upward parabola is positive outside the roots, and the inclusive ≥\ge≥ keeps them, so x≤3−5x \le 3 - \sqrt{5}x≤3−5 or x≥3+5x \ge 3 + \sqrt{5}x≥3+5.
Solve 9−x2≤09 - x^2 \le 09−x2≤0.
Multiply by −1-1−1 and flip the symbol.
9−x2≤0⇒x2−9=(x−3)(x+3)≥09 - x^2 \le 0 \Rightarrow x^2 - 9 = (x - 3)(x + 3) \ge 09−x2≤0⇒x2−9=(x−3)(x+3)≥0
The upward parabola is positive outside the roots −3-3−3 and 333, and the inclusive symbol keeps them, so x≤−3x \le -3x≤−3 or x≥3x \ge 3x≥3.
For which xxx is the parabola y=x2−4xy = x^2 - 4xy=x2−4x below the x-axis?
Below the x-axis means y<0y < 0y<0. Factor to find the roots.
x2−4x=x(x−4)=0⇒x=0 or x=4x^2 - 4x = x(x - 4) = 0 \Rightarrow x = 0 \text{ or } x = 4x2−4x=x(x−4)=0⇒x=0 or x=4
The upward parabola is below the axis between the roots, so 0<x<40 < x < 40<x<4.
Solve x2+8x+16≥0x^2 + 8x + 16 \ge 0x2+8x+16≥0.
x2+8x+16=(x+4)2≥0 for every xx^2 + 8x + 16 = (x + 4)^2 \ge 0 \text{ for every } xx2+8x+16=(x+4)2≥0 for every x
A square is never negative, so (x+4)2≥0(x + 4)^2 \ge 0(x+4)2≥0 is true for all real numbers.
Solve 3x2+5x−2>03x^2 + 5x - 2 > 03x2+5x−2>0.
Factor the quadratic with a leading coefficient of 333.
3x2+5x−2=(3x−1)(x+2)=0⇒x=−2 or x=133x^2 + 5x - 2 = (3x - 1)(x + 2) = 0 \Rightarrow x = -2 \text{ or } x = \tfrac{1}{3}3x2+5x−2=(3x−1)(x+2)=0⇒x=−2 or x=31
The upward parabola is positive outside the roots, and the strict symbol excludes them, so x<−2x < -2x<−2 or x>13x > \tfrac{1}{3}x>31.
Solve x2−2x−1>0x^2 - 2x - 1 > 0x2−2x−1>0.
Use the quadratic formula with a=1a = 1a=1, b=−2b = -2b=−2, c=−1c = -1c=−1.
x=2±4+42=1±2x = \frac{2 \pm \sqrt{4 + 4}}{2} = 1 \pm \sqrt{2}x=22±4+4=1±2
The upward parabola is positive outside the roots, and the strict symbol excludes them, so x<1−2x < 1 - \sqrt{2}x<1−2 or x>1+2x > 1 + \sqrt{2}x>1+2.
Reset this practice set?
This clears every answer you have given and starts the set again from question 1.