12 multiple-choice questions, progressively harder.
For the roots of 3x2−15x+73x^2 - 15x + 73x2−15x+7, find r12+r22r_1^2 + r_2^2r12+r22.
Solution
Correct answer: D
The sum is −ba=5-\frac{b}{a} = 5−ab=5 and the product is ca=73\frac{c}{a} = \frac{7}{3}ac=37. Apply r12+r22=(r1+r2)2−2r1r2r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2 r_1 r_2r12+r22=(r1+r2)2−2r1r2.
r12+r22=52−2(73)=25−143=613r_1^2 + r_2^2 = 5^2 - 2\left(\tfrac{7}{3}\right) = 25 - \tfrac{14}{3} = \tfrac{61}{3}r12+r22=52−2(37)=25−314=361
Subtract twice the product from the square of the sum.
For the roots of 3x2−12x+53x^2 - 12x + 53x2−12x+5, find (r1−r2)2(r_1 - r_2)^2(r1−r2)2.
Correct answer: B
Use (r1−r2)2=Δa2(r_1 - r_2)^2 = \frac{\Delta}{a^2}(r1−r2)2=a2Δ with Δ=(−12)2−4(3)(5)=84\Delta = (-12)^2 - 4(3)(5) = 84Δ=(−12)2−4(3)(5)=84 and a=3a = 3a=3.
(r1−r2)2=849=283(r_1 - r_2)^2 = \frac{84}{9} = \frac{28}{3}(r1−r2)2=984=328
The squared gap is the discriminant over a2a^2a2.
For the roots of x2−2x−5x^2 - 2x - 5x2−2x−5, find r13+r23r_1^3 + r_2^3r13+r23.
Correct answer: C
Use r13+r23=(r1+r2)3−3r1r2(r1+r2)r_1^3 + r_2^3 = (r_1 + r_2)^3 - 3 r_1 r_2 (r_1 + r_2)r13+r23=(r1+r2)3−3r1r2(r1+r2) with sum 222 and product −5-5−5.
r13+r23=23−3(−5)(2)=8+30=38r_1^3 + r_2^3 = 2^3 - 3(-5)(2) = 8 + 30 = 38r13+r23=23−3(−5)(2)=8+30=38
The negative product adds to the total.
One root of 2x2+bx−6=02x^2 + bx - 6 = 02x2+bx−6=0 is −3-3−3. Find the other root and bbb.
The roots multiply to ca=−62=−3\frac{c}{a} = \frac{-6}{2} = -3ac=2−6=−3, so the other root is −3−3=1\frac{-3}{-3} = 1−3−3=1. The roots then sum to −3+1=−2-3 + 1 = -2−3+1=−2, which equals −ba=−b2-\frac{b}{a} = -\frac{b}{2}−ab=−2b.
−b2=−2 ⇒ b=4-\frac{b}{2} = -2 \;\Rightarrow\; b = 4−2b=−2⇒b=4
The other root is 111 and b=4b = 4b=4.
Do two real numbers with sum 888 and product 151515 exist?
Test the condition s2−4p≥0s^2 - 4p \ge 0s2−4p≥0 with s=8s = 8s=8 and p=15p = 15p=15.
82−4(15)=64−60=4≥08^2 - 4(15) = 64 - 60 = 4 \ge 082−4(15)=64−60=4≥0
The condition holds, and the numbers are the roots of x2−8x+15=(x−3)(x−5)x^2 - 8x + 15 = (x - 3)(x - 5)x2−8x+15=(x−3)(x−5), namely 333 and 555.
Do two real numbers with sum 222 and product 333 exist?
Test the condition s2−4p≥0s^2 - 4p \ge 0s2−4p≥0 with s=2s = 2s=2 and p=3p = 3p=3.
22−4(3)=4−12=−8<02^2 - 4(3) = 4 - 12 = -8 < 022−4(3)=4−12=−8<0
The condition fails, so no real numbers have sum 222 and product 333.
Which of the following is NOT symmetric in the roots r1r_1r1 and r2r_2r2 (unchanged when they are swapped)?
Correct answer: A
Swapping r1r_1r1 and r2r_2r2 leaves a symmetric expression unchanged. The sum, product, and sum of squares are all symmetric, but the difference flips sign.
r1−r2 → r2−r1=−(r1−r2)r_1 - r_2 \;\to\; r_2 - r_1 = -(r_1 - r_2)r1−r2→r2−r1=−(r1−r2)
Since it changes, r1−r2r_1 - r_2r1−r2 is not symmetric, and so is not determined by the coefficients.
Build a monic quadratic whose roots are each 333 more than the roots of x2−4x+3x^2 - 4x + 3x2−4x+3.
The roots of x2−4x+3x^2 - 4x + 3x2−4x+3 have sum 444 and product 333. The new roots r1+3r_1 + 3r1+3 and r2+3r_2 + 3r2+3 have sum (r1+r2)+6=10(r_1 + r_2) + 6 = 10(r1+r2)+6=10 and product r1r2+3(r1+r2)+9=3+12+9=24r_1 r_2 + 3(r_1 + r_2) + 9 = 3 + 12 + 9 = 24r1r2+3(r1+r2)+9=3+12+9=24.
x2−10x+24x^2 - 10x + 24x2−10x+24
The shifted sum and product feed the monic template.
For 4x2−12x+c4x^2 - 12x + c4x2−12x+c to have real roots, which condition must ccc satisfy?
Real roots require Δ=(−12)2−4(4)c≥0\Delta = (-12)^2 - 4(4)c \ge 0Δ=(−12)2−4(4)c≥0.
144−16c≥0 ⇒ c≤9144 - 16c \ge 0 \;\Rightarrow\; c \le 9144−16c≥0⇒c≤9
So ccc can be at most 999.
The quadratic x2+bx+12x^2 + bx + 12x2+bx+12 has one root that is three times the other, and both roots are positive. Find bbb.
Let the roots be rrr and 3r3r3r. Their product is the constant 121212.
r⋅3r=3r2=12 ⇒ r=2r \cdot 3r = 3r^2 = 12 \;\Rightarrow\; r = 2r⋅3r=3r2=12⇒r=2
The roots are 222 and 666, so their sum is 888. Since the sum is −b-b−b, we get b=−8b = -8b=−8.
For the roots of x2−8x+4x^2 - 8x + 4x2−8x+4, find r1r2+r2r1\dfrac{r_1}{r_2} + \dfrac{r_2}{r_1}r2r1+r1r2.
Combine over a common denominator: r1r2+r2r1=r12+r22r1r2\frac{r_1}{r_2} + \frac{r_2}{r_1} = \frac{r_1^2 + r_2^2}{r_1 r_2}r2r1+r1r2=r1r2r12+r22. With sum 888 and product 444, the numerator is 82−2(4)=568^2 - 2(4) = 5682−2(4)=56.
r1r2+r2r1=564=14\frac{r_1}{r_2} + \frac{r_2}{r_1} = \frac{56}{4} = 14r2r1+r1r2=456=14
The expression reduces to the sum and product.
Find the sum of the roots of 6x2+7x−36x^2 + 7x - 36x2+7x−3.
The sum of the roots is −ba-\frac{b}{a}−ab with a=6a = 6a=6 and b=7b = 7b=7.
r1+r2=−76r_1 + r_2 = -\frac{7}{6}r1+r2=−67
The minus sign lands on the sum.
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