12 multiple-choice questions, progressively harder.
Find the sum of the roots of x2−7x+12x^2 - 7x + 12x2−7x+12.
Solution
Correct answer: C
For a monic quadratic x2+bx+cx^2 + bx + cx2+bx+c, the sum of the roots is −b-b−b. Here b=−7b = -7b=−7.
r1+r2=−b=−(−7)=7r_1 + r_2 = -b = -(-7) = 7r1+r2=−b=−(−7)=7
The minus sign in front of bbb turns the −7-7−7 into +7+7+7.
Find the product of the roots of x2−7x+12x^2 - 7x + 12x2−7x+12.
Correct answer: A
For a monic quadratic x2+bx+cx^2 + bx + cx2+bx+c, the product of the roots is ccc. Here c=12c = 12c=12.
r1r2=c=12r_1 r_2 = c = 12r1r2=c=12
The product keeps the sign of the constant term, with no extra minus.
Find the sum of the roots of x2+5x+6x^2 + 5x + 6x2+5x+6.
Correct answer: D
The sum of the roots of a monic quadratic is −b-b−b, and here b=5b = 5b=5.
r1+r2=−b=−5r_1 + r_2 = -b = -5r1+r2=−b=−5
The roots (which are −2-2−2 and −3-3−3) add to −5-5−5, matching the sign of −b-b−b.
For a quadratic ax2+bx+cax^2 + bx + cax2+bx+c with a≠0a \ne 0a=0, the sum of the roots equals which expression?
Correct answer: B
Expanding a(x−r1)(x−r2)=ax2−a(r1+r2)x+a r1r2a(x - r_1)(x - r_2) = ax^2 - a(r_1 + r_2)x + a\,r_1 r_2a(x−r1)(x−r2)=ax2−a(r1+r2)x+ar1r2 and matching the middle coefficient gives b=−a(r1+r2)b = -a(r_1 + r_2)b=−a(r1+r2).
r1+r2=−bar_1 + r_2 = -\frac{b}{a}r1+r2=−ab
Only the sum carries the minus sign.
Find the sum of the roots of 2x2−8x+62x^2 - 8x + 62x2−8x+6.
With a leading coefficient, the sum of the roots is −ba-\frac{b}{a}−ab. Here a=2a = 2a=2 and b=−8b = -8b=−8.
r1+r2=−ba=−−82=4r_1 + r_2 = -\frac{b}{a} = -\frac{-8}{2} = 4r1+r2=−ab=−2−8=4
Divide by aaa, then apply the minus sign.
Which monic quadratic has roots 222 and 555?
Find the sum and product of the roots, then use the template x2−(r1+r2)x+r1r2x^2 - (r_1 + r_2)x + r_1 r_2x2−(r1+r2)x+r1r2.
r1+r2=7,r1r2=10 ⇒ x2−7x+10r_1 + r_2 = 7, \qquad r_1 r_2 = 10 \;\Rightarrow\; x^2 - 7x + 10r1+r2=7,r1r2=10⇒x2−7x+10
The middle coefficient is the negative of the sum.
One root of x2−10x+16x^2 - 10x + 16x2−10x+16 is 222. What is the other root?
The roots sum to −b=10-b = 10−b=10. Subtract the known root from that total.
r2=(r1+r2)−r1=10−2=8r_2 = (r_1 + r_2) - r_1 = 10 - 2 = 8r2=(r1+r2)−r1=10−2=8
As a check, the product is 161616 and 2⋅8=162 \cdot 8 = 162⋅8=16.
One root of x2+2x−15x^2 + 2x - 15x2+2x−15 is 333. What is the other root?
The roots sum to −b=−2-b = -2−b=−2. Subtract the known root.
r2=(r1+r2)−r1=−2−3=−5r_2 = (r_1 + r_2) - r_1 = -2 - 3 = -5r2=(r1+r2)−r1=−2−3=−5
The product confirms it: 3⋅(−5)=−15=c3 \cdot (-5) = -15 = c3⋅(−5)=−15=c.
A monic quadratic has roots that sum to 444 and multiply to 333. Which quadratic is it?
Use the template x2−(sum)x+(product)x^2 - (\text{sum})x + (\text{product})x2−(sum)x+(product).
x2−4x+3x^2 - 4x + 3x2−4x+3
The middle coefficient is the negative of the sum 444, and the constant is the product 333.
Find the product of the roots of x2−6x+9x^2 - 6x + 9x2−6x+9.
The product of the roots is c=9c = 9c=9.
r1r2=c=9r_1 r_2 = c = 9r1r2=c=9
This quadratic has the repeated root 333, and 3⋅3=93 \cdot 3 = 93⋅3=9.
Find the sum of the roots of 4x2−12x+54x^2 - 12x + 54x2−12x+5.
The sum of the roots is −ba-\frac{b}{a}−ab with a=4a = 4a=4 and b=−12b = -12b=−12.
r1+r2=−−124=3r_1 + r_2 = -\frac{-12}{4} = 3r1+r2=−4−12=3
Divide by the leading coefficient, then apply the minus sign.
Find the product of the roots of 5x2−20x+155x^2 - 20x + 155x2−20x+15.
This quadratic factors as 5(x−1)(x−3)5(x - 1)(x - 3)5(x−1)(x−3), so its roots are the real numbers 111 and 333. The product of the roots is ca\frac{c}{a}ac with a=5a = 5a=5 and c=15c = 15c=15.
r1r2=ca=155=3r_1 r_2 = \frac{c}{a} = \frac{15}{5} = 3r1r2=ac=515=3
As a check, 1⋅3=31 \cdot 3 = 31⋅3=3, matching the constant divided by the leading coefficient.
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