12 multiple-choice questions, progressively harder.
Find the sum of the roots of x2+8x+15x^2 + 8x + 15x2+8x+15.
Solution
Correct answer: B
The roots of x2+bx+cx^2 + bx + cx2+bx+c add to −b-b−b. Here b=8b = 8b=8.
r+s=−b=−8r + s = -b = -8r+s=−b=−8
So the sum of the roots is −8-8−8. The value 888 drops the minus sign that always sits in front of bbb.
Which monic quadratic equation has roots −1-1−1 and 444?
Correct answer: A
The sum is −1+4=3-1 + 4 = 3−1+4=3 and the product is (−1)(4)=−4(-1)(4) = -4(−1)(4)=−4. Use x2−(r+s)x+rsx^2 - (r + s)x + rsx2−(r+s)x+rs.
x2−3x−4=0x^2 - 3x - 4 = 0x2−3x−4=0
The negative of the sum 333 is the middle coefficient −3-3−3. The choice x2+3x−4x^2 + 3x - 4x2+3x−4 misses that flip.
One root of x2−9x+20=0x^2 - 9x + 20 = 0x2−9x+20=0 is 444. Find the other root.
The two roots add to −b=9-b = 9−b=9. Subtract the known root from that total.
s=(r+s)−r=9−4=5s = (r + s) - r = 9 - 4 = 5s=(r+s)−r=9−4=5
So the other root is 555. The product agrees, since 20÷4=520 \div 4 = 520÷4=5.
Find the sum of the roots of x2−x−6x^2 - x - 6x2−x−6.
The roots of x2+bx+cx^2 + bx + cx2+bx+c add to −b-b−b. Here b=−1b = -1b=−1.
r+s=−b=−(−1)=1r + s = -b = -(-1) = 1r+s=−b=−(−1)=1
So the sum of the roots is 111. The value −1-1−1 forgets the double negative.
Find the product of the roots of x2−x−6x^2 - x - 6x2−x−6.
Correct answer: C
The roots multiply to ccc. Here c=−6c = -6c=−6.
rs=c=−6rs = c = -6rs=c=−6
So the product of the roots is −6-6−6. The value −1-1−1 is the sum, not the product.
For the monic quadratic x2+bx+cx^2 + bx + cx2+bx+c with roots rrr and sss, which statement is correct?
Correct answer: D
Match x2+bx+cx^2 + bx + cx2+bx+c with (x−r)(x−s)=x2−(r+s)x+rs(x - r)(x - s) = x^2 - (r + s)x + rs(x−r)(x−s)=x2−(r+s)x+rs term by term. The middle terms give b=−(r+s)b = -(r + s)b=−(r+s) and the constants give c=rsc = rsc=rs.
r+s=−b,rs=cr + s = -b, \qquad rs = cr+s=−b,rs=c
Only the sum picks up the minus sign; the product equals ccc directly.
One root of x2−10x+21=0x^2 - 10x + 21 = 0x2−10x+21=0 is 777. Find the other root.
The two roots add to −b=10-b = 10−b=10. Subtract the known root.
s=10−7=3s = 10 - 7 = 3s=10−7=3
So the other root is 333. The product checks it, since 21÷7=321 \div 7 = 321÷7=3.
Find the product of the roots of x2+7x+12x^2 + 7x + 12x2+7x+12.
The roots multiply to ccc. Here c=12c = 12c=12.
rs=c=12rs = c = 12rs=c=12
So the product of the roots is 121212. The value −7-7−7 is the sum, not the product.
Find the sum of the roots of x2+2x−15x^2 + 2x - 15x2+2x−15.
The roots of x2+bx+cx^2 + bx + cx2+bx+c add to −b-b−b. Here b=2b = 2b=2.
r+s=−b=−2r + s = -b = -2r+s=−b=−2
So the sum of the roots is −2-2−2. The value 222 drops the minus sign in front of bbb.
Which monic quadratic equation has roots −3-3−3 and −5-5−5?
The sum is −3+(−5)=−8-3 + (-5) = -8−3+(−5)=−8 and the product is (−3)(−5)=15(-3)(-5) = 15(−3)(−5)=15. Use x2−(r+s)x+rsx^2 - (r + s)x + rsx2−(r+s)x+rs.
x2−(−8)x+15=x2+8x+15=0x^2 - (-8)x + 15 = x^2 + 8x + 15 = 0x2−(−8)x+15=x2+8x+15=0
The negative of the sum −8-8−8 is the positive middle coefficient +8+8+8.
Find the sum and product of the roots of x2−6x+8x^2 - 6x + 8x2−6x+8.
Here b=−6b = -6b=−6 and c=8c = 8c=8. The sum is −b-b−b and the product is ccc.
r+s=−b=6,rs=c=8r + s = -b = 6, \qquad rs = c = 8r+s=−b=6,rs=c=8
So the sum is 666 and the product is 888. The choice with sum −6-6−6 forgets the sign flip.
A monic quadratic x2+bx+cx^2 + bx + cx2+bx+c has roots 333 and 444. What is bbb?
The roots add to −b-b−b, so bbb is the negative of the sum of the roots.
b=−(r+s)=−(3+4)=−7b = -(r + s) = -(3 + 4) = -7b=−(r+s)=−(3+4)=−7
So b=−7b = -7b=−7. The value 121212 is the product ccc, not bbb.
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