12 multiple-choice questions, progressively harder.
Find the product of the roots of 3x2+7x−23x^2 + 7x - 23x2+7x−2.
Solution
Correct answer: A
The product of the roots is ca\frac{c}{a}ac with a=3a = 3a=3 and c=−2c = -2c=−2.
r1r2=−23=−23r_1 r_2 = \frac{-2}{3} = -\frac{2}{3}r1r2=3−2=−32
The product keeps the constant's sign.
Find the sum of the roots of 5x2+3x−15x^2 + 3x - 15x2+3x−1.
Correct answer: C
The sum of the roots is −ba-\frac{b}{a}−ab with a=5a = 5a=5 and b=3b = 3b=3.
r1+r2=−35r_1 + r_2 = -\frac{3}{5}r1+r2=−53
The minus sign lands on the sum.
One root of 2x2−7x+3=02x^2 - 7x + 3 = 02x2−7x+3=0 is 333. What is the other root?
The roots multiply to ca=32\frac{c}{a} = \frac{3}{2}ac=23. Divide by the known root.
r2=r1r2r1=3/23=12r_2 = \frac{r_1 r_2}{r_1} = \frac{3/2}{3} = \frac{1}{2}r2=r1r1r2=33/2=21
The sum confirms it: 3+12=72=−ba3 + \tfrac{1}{2} = \tfrac{7}{2} = -\frac{b}{a}3+21=27=−ab.
One root of 3x2−10x+3=03x^2 - 10x + 3 = 03x2−10x+3=0 is 333. What is the other root?
Correct answer: D
The roots multiply to ca=33=1\frac{c}{a} = \frac{3}{3} = 1ac=33=1. Divide by the known root.
r2=13r_2 = \frac{1}{3}r2=31
The sum checks: 3+13=103=−ba3 + \tfrac{1}{3} = \tfrac{10}{3} = -\frac{b}{a}3+31=310=−ab.
Which quadratic with leading coefficient 222 has roots 12\tfrac{1}{2}21 and 333?
The sum is 12+3=72\tfrac{1}{2} + 3 = \tfrac{7}{2}21+3=27 and the product is 12⋅3=32\tfrac{1}{2} \cdot 3 = \tfrac{3}{2}21⋅3=23, giving the monic x2−72x+32x^2 - \tfrac{7}{2}x + \tfrac{3}{2}x2−27x+23. Multiply through by 222.
2(x2−72x+32)=2x2−7x+32\left(x^2 - \tfrac{7}{2}x + \tfrac{3}{2}\right) = 2x^2 - 7x + 32(x2−27x+23)=2x2−7x+3
Scaling by 222 clears the fractions without moving the roots.
For the roots of x2−5x+3x^2 - 5x + 3x2−5x+3, find r12+r22r_1^2 + r_2^2r12+r22.
Correct answer: B
The quadratic is monic, so r1+r2=5r_1 + r_2 = 5r1+r2=5 and r1r2=3r_1 r_2 = 3r1r2=3. Use 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(3)=25−6=19r_1^2 + r_2^2 = 5^2 - 2(3) = 25 - 6 = 19r12+r22=52−2(3)=25−6=19
Subtract twice the product from the square of the sum.
For the roots of x2−7x+3x^2 - 7x + 3x2−7x+3, find 1r1+1r2\dfrac{1}{r_1} + \dfrac{1}{r_2}r11+r21.
Combine the reciprocals over a common denominator: 1r1+1r2=r1+r2r1r2\frac{1}{r_1} + \frac{1}{r_2} = \frac{r_1 + r_2}{r_1 r_2}r11+r21=r1r2r1+r2. Here the sum is 777 and the product is 333.
1r1+1r2=73\frac{1}{r_1} + \frac{1}{r_2} = \frac{7}{3}r11+r21=37
The reciprocals depend only on the sum and product.
Do two real numbers with sum 333 and product 555 exist?
Test the condition s2−4p≥0s^2 - 4p \ge 0s2−4p≥0 with s=3s = 3s=3 and p=5p = 5p=5.
32−4(5)=9−20=−11<03^2 - 4(5) = 9 - 20 = -11 < 032−4(5)=9−20=−11<0
The condition fails, so no real numbers have that sum and product.
For the roots of 2x2−8x+32x^2 - 8x + 32x2−8x+3, find r12+r22r_1^2 + r_2^2r12+r22.
The sum is −ba=4-\frac{b}{a} = 4−ab=4 and the product is ca=32\frac{c}{a} = \frac{3}{2}ac=23. 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=42−2(32)=16−3=13r_1^2 + r_2^2 = 4^2 - 2\left(\tfrac{3}{2}\right) = 16 - 3 = 13r12+r22=42−2(23)=16−3=13
The cross term is twice the product.
For the roots of x2−6x+4x^2 - 6x + 4x2−6x+4, find (r1−r2)2(r_1 - r_2)^2(r1−r2)2.
Use (r1−r2)2=(r1+r2)2−4r1r2(r_1 - r_2)^2 = (r_1 + r_2)^2 - 4 r_1 r_2(r1−r2)2=(r1+r2)2−4r1r2 with sum 666 and product 444.
(r1−r2)2=62−4(4)=36−16=20(r_1 - r_2)^2 = 6^2 - 4(4) = 36 - 16 = 20(r1−r2)2=62−4(4)=36−16=20
This equals the discriminant, since the quadratic is monic.
One root of x2+x−12=0x^2 + x - 12 = 0x2+x−12=0 is 333. What is the other root?
The roots sum to −b=−1-b = -1−b=−1. Subtract the known root.
r2=−1−3=−4r_2 = -1 - 3 = -4r2=−1−3=−4
The product confirms it: 3⋅(−4)=−12=c3 \cdot (-4) = -12 = c3⋅(−4)=−12=c.
Which quadratic with leading coefficient 222 has roots −12-\tfrac{1}{2}−21 and −3-3−3?
The sum is −12+(−3)=−72-\tfrac{1}{2} + (-3) = -\tfrac{7}{2}−21+(−3)=−27 and the product is −12⋅(−3)=32-\tfrac{1}{2} \cdot (-3) = \tfrac{3}{2}−21⋅(−3)=23, so the monic quadratic is x2+72x+32x^2 + \tfrac{7}{2}x + \tfrac{3}{2}x2+27x+23. Multiply by 222.
2(x2+72x+32)=2x2+7x+32\left(x^2 + \tfrac{7}{2}x + \tfrac{3}{2}\right) = 2x^2 + 7x + 32(x2+27x+23)=2x2+7x+3
Factoring back to (2x+1)(x+3)(2x + 1)(x + 3)(2x+1)(x+3) confirms the roots.
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