12 multiple-choice questions, progressively harder.
Both x=−2x = -2x=−2 and x=6x = 6x=6 are proposed as solutions of x2−4x−12=0x^2 - 4x - 12 = 0x2−4x−12=0. Which are actually correct?
Solution
Correct answer: B
Substitute each candidate into x2−4x−12x^2 - 4x - 12x2−4x−12.
(−2)2−4(−2)−12=4+8−12=0,62−4(6)−12=36−24−12=0(-2)^2 - 4(-2) - 12 = 4 + 8 - 12 = 0, \qquad 6^2 - 4(6) - 12 = 36 - 24 - 12 = 0(−2)2−4(−2)−12=4+8−12=0,62−4(6)−12=36−24−12=0
Both give 000, so both are solutions. A quadratic can have two roots, and here it does.
Solve 5x2=455x^2 = 455x2=45.
Divide both sides by 555 to isolate the square, then take roots.
x2=9 ⇒ x=±3x^2 = 9 \;\Rightarrow\; x = \pm 3x2=9⇒x=±3
So x=3x = 3x=3 or x=−3x = -3x=−3.
Solve x2=94x^2 = \tfrac{9}{4}x2=49.
Correct answer: C
Take the square root of both sides; the root of a fraction is the root of the top over the root of the bottom.
x=±94=±32x = \pm\sqrt{\tfrac{9}{4}} = \pm\tfrac{3}{2}x=±49=±23
So x=32x = \tfrac{3}{2}x=23 or x=−32x = -\tfrac{3}{2}x=−23.
Which value is a solution of x2+3x−10=0x^2 + 3x - 10 = 0x2+3x−10=0?
Substitute the candidates into x2+3x−10x^2 + 3x - 10x2+3x−10 and look for 000.
22+3(2)−10=4+6−10=02^2 + 3(2) - 10 = 4 + 6 - 10 = 022+3(2)−10=4+6−10=0
So x=2x = 2x=2 is a solution. (The other root, −5-5−5, is not listed.)
A square has area 505050 square units, so its side length sss satisfies s2=50s^2 = 50s2=50. What is sss?
Correct answer: D
Solve s2=50s^2 = 50s2=50 by square roots, then keep only the meaningful root for a length.
s=50=52s = \sqrt{50} = 5\sqrt{2}s=50=52
Algebraically s=±52s = \pm 5\sqrt{2}s=±52, but a side length cannot be negative, so s=52s = 5\sqrt{2}s=52.
Solve x2=19x^2 = \tfrac{1}{9}x2=91.
Take the square root of both sides.
x=±19=±13x = \pm\sqrt{\tfrac{1}{9}} = \pm\tfrac{1}{3}x=±91=±31
So x=13x = \tfrac{1}{3}x=31 or x=−13x = -\tfrac{1}{3}x=−31.
Expanding (x−3)(x+3)(x - 3)(x + 3)(x−3)(x+3) gives which standard form?
Correct answer: A
Multiply out; the middle terms −3x-3x−3x and +3x+3x+3x cancel.
(x−3)(x+3)=x2+3x−3x−9=x2−9(x - 3)(x + 3) = x^2 + 3x - 3x - 9 = x^2 - 9(x−3)(x+3)=x2+3x−3x−9=x2−9
A quadratic has roots x=0x = 0x=0 and x=8x = 8x=8. In factored form it is:
The root 000 gives the factor xxx, and the root 888 gives the factor x−8x - 8x−8.
x(x−8)=0x(x - 8) = 0x(x−8)=0
Check: x=0x = 0x=0 zeros the first factor, and x=8x = 8x=8 zeros the second.
To solve (x−3)(x−1)=3(x - 3)(x - 1) = 3(x−3)(x−1)=3 with the zero-product property, what must you do first?
The zero-product property only works when the product is 000, but right now the product equals 333.
(x−3)(x−1)=3 must become …=0(x - 3)(x - 1) = 3 \ \text{must become}\ \ldots = 0(x−3)(x−1)=3 must become …=0
Expand and move the 333 across to reach standard form first, then solve. You cannot set a factor equal to 333.
Solve 2(x−1)2=82(x - 1)^2 = 82(x−1)2=8.
Divide both sides by 222 to isolate the squared group.
(x−1)2=4 ⇒ x−1=±2(x - 1)^2 = 4 \;\Rightarrow\; x - 1 = \pm 2(x−1)2=4⇒x−1=±2
Then x=1+2=3x = 1 + 2 = 3x=1+2=3 or x=1−2=−1x = 1 - 2 = -1x=1−2=−1.
Solve (x−9)(x+9)=0(x - 9)(x + 9) = 0(x−9)(x+9)=0.
Set each factor equal to 000.
x−9=0orx+9=0x - 9 = 0 \quad \text{or} \quad x + 9 = 0x−9=0orx+9=0
So x=9x = 9x=9 or x=−9x = -9x=−9.
Which equation has two real solutions?
Two solutions need a positive number on the right after the square is isolated.
x2=16 ⇒ x=±4x^2 = 16 \;\Rightarrow\; x = \pm 4x2=16⇒x=±4
The others give one solution (x2=0x^2 = 0x2=0 and (x−4)2=0(x - 4)^2 = 0(x−4)2=0) or none (x2=−16x^2 = -16x2=−16).
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