12 multiple-choice questions, progressively harder.
Find the sum of the roots of 5x2−11x+25x^2 - 11x + 25x2−11x+2.
Solution
Correct answer: A
Here a=5a = 5a=5 and b=−11b = -11b=−11, and the roots add to −ba-\frac{b}{a}−ab.
r+s=−−115=115r + s = -\frac{-11}{5} = \frac{11}{5}r+s=−5−11=511
So the sum is 115\tfrac{11}{5}511. The value −115-\tfrac{11}{5}−511 forgets that bbb already carries a minus.
Which monic quadratic equation has roots 5+225 + 2\sqrt{2}5+22 and 5−225 - 2\sqrt{2}5−22?
Correct answer: C
The sum is (5+22)+(5−22)=10(5 + 2\sqrt{2}) + (5 - 2\sqrt{2}) = 10(5+22)+(5−22)=10, and the product is a difference of squares.
rs=(5)2−(22)2=25−8=17rs = (5)^2 - (2\sqrt{2})^2 = 25 - 8 = 17rs=(5)2−(22)2=25−8=17
Using x2−(r+s)x+rsx^2 - (r + s)x + rsx2−(r+s)x+rs gives x2−10x+17=0x^2 - 10x + 17 = 0x2−10x+17=0.
The roots of x2−kx+12=0x^2 - kx + 12 = 0x2−kx+12=0 are 333 and 444. Find kkk.
Correct answer: D
The sum of the roots is −(−k)=k-(-k) = k−(−k)=k. So kkk equals the sum of the roots.
k=r+s=3+4=7k = r + s = 3 + 4 = 7k=r+s=3+4=7
So k=7k = 7k=7. The product 121212 already matches ccc, which confirms the setup.
Which quadratic with integer coefficients and leading coefficient 444 has roots 34\tfrac{3}{4}43 and −2-2−2?
The roots sum to 34+(−2)=−54\tfrac{3}{4} + (-2) = -\tfrac{5}{4}43+(−2)=−45 and multiply to 34⋅(−2)=−32\tfrac{3}{4} \cdot (-2) = -\tfrac{3}{2}43⋅(−2)=−23, so the monic form is x2+54x−32x^2 + \tfrac{5}{4}x - \tfrac{3}{2}x2+45x−23. Multiply through by 444.
4(x2+54x−32)=4x2+5x−6=04\left(x^2 + \tfrac{5}{4}x - \tfrac{3}{2}\right) = 4x^2 + 5x - 6 = 04(x2+45x−23)=4x2+5x−6=0
This factors as (4x−3)(x+2)(4x - 3)(x + 2)(4x−3)(x+2), confirming the roots.
A student claims 2+32 + \sqrt{3}2+3 and 2−32 - \sqrt{3}2−3 are the roots of x2−4x+1=0x^2 - 4x + 1 = 0x2−4x+1=0. Checking with the sum and product, is the claim correct?
Correct answer: B
The roots must sum to −b=4-b = 4−b=4 and multiply to c=1c = 1c=1. Test the pair against both.
(2+3)+(2−3)=4,(2+3)(2−3)=4−3=1(2 + \sqrt{3}) + (2 - \sqrt{3}) = 4, \qquad (2 + \sqrt{3})(2 - \sqrt{3}) = 4 - 3 = 1(2+3)+(2−3)=4,(2+3)(2−3)=4−3=1
Both totals match, so the claim is correct.
Find the product of the roots of 3x2+12=11x3x^2 + 12 = 11x3x2+12=11x.
In standard form the equation is 3x2−11x+12=03x^2 - 11x + 12 = 03x2−11x+12=0, so a=3a = 3a=3 and c=12c = 12c=12.
rs=ca=123=4rs = \frac{c}{a} = \frac{12}{3} = 4rs=ac=312=4
So the product is 444.
One root of 2x2−9x+c=02x^2 - 9x + c = 02x2−9x+c=0 is 12\tfrac{1}{2}21. Find ccc.
The roots add to −ba=92-\frac{b}{a} = \frac{9}{2}−ab=29, so the other root is 92−12=4\frac{9}{2} - \frac{1}{2} = 429−21=4. The product of the roots is ca=c2\frac{c}{a} = \frac{c}{2}ac=2c.
12⋅4=c2 ⇒ 2=c2 ⇒ c=4\frac{1}{2} \cdot 4 = \frac{c}{2} \ \Rightarrow \ 2 = \frac{c}{2} \ \Rightarrow \ c = 421⋅4=2c ⇒ 2=2c ⇒ c=4
So c=4c = 4c=4.
The roots of x2+bx−21=0x^2 + bx - 21 = 0x2+bx−21=0 are 333 and −7-7−7. Find bbb.
The middle coefficient bbb is the negative of the sum of the roots.
b=−(r+s)=−(3+(−7))=−(−4)=4b = -(r + s) = -(3 + (-7)) = -(-4) = 4b=−(r+s)=−(3+(−7))=−(−4)=4
So b=4b = 4b=4. The product 3⋅(−7)=−213 \cdot (-7) = -213⋅(−7)=−21 matches ccc, confirming the setup.
Write a quadratic with leading coefficient 222 whose roots are −3-3−3 and 12\tfrac{1}{2}21.
The roots sum to −3+12=−52-3 + \tfrac{1}{2} = -\tfrac{5}{2}−3+21=−25 and multiply to (−3)(12)=−32(-3)\left(\tfrac{1}{2}\right) = -\tfrac{3}{2}(−3)(21)=−23, so the monic form is x2+52x−32x^2 + \tfrac{5}{2}x - \tfrac{3}{2}x2+25x−23. Multiply through by 222.
2(x2+52x−32)=2x2+5x−3=02\left(x^2 + \tfrac{5}{2}x - \tfrac{3}{2}\right) = 2x^2 + 5x - 3 = 02(x2+25x−23)=2x2+5x−3=0
This is also 2(x+3)(x−12)=(x+3)(2x−1)2(x + 3)\left(x - \tfrac{1}{2}\right) = (x + 3)(2x - 1)2(x+3)(x−21)=(x+3)(2x−1), with the required roots.
Two roots of a quadratic sum to −12-\tfrac{1}{2}−21, and one root is 32\tfrac{3}{2}23. Find the other root.
The other root is the total sum minus the known root.
s=−12−32=−2s = -\frac{1}{2} - \frac{3}{2} = -2s=−21−23=−2
So the other root is −2-2−2.
A monic quadratic x2+bx+cx^2 + bx + cx2+bx+c has roots whose sum is 000. What must be true of its coefficients?
The sum of the roots is −b-b−b. If that sum is 000, then −b=0-b = 0−b=0.
−b=0 ⇒ b=0-b = 0 \ \Rightarrow \ b = 0−b=0 ⇒ b=0
So the middle term vanishes. For example x2−9x^2 - 9x2−9 has roots 333 and −3-3−3, which sum to 000.
Find a monic quadratic equation whose roots are 1+61 + \sqrt{6}1+6 and 1−61 - \sqrt{6}1−6.
The sum is (1+6)+(1−6)=2(1 + \sqrt{6}) + (1 - \sqrt{6}) = 2(1+6)+(1−6)=2, and the product is a difference of squares.
rs=(1)2−(6)2=1−6=−5rs = (1)^2 - (\sqrt{6})^2 = 1 - 6 = -5rs=(1)2−(6)2=1−6=−5
Using x2−(r+s)x+rsx^2 - (r + s)x + rsx2−(r+s)x+rs gives x2−2x−5=0x^2 - 2x - 5 = 0x2−2x−5=0.
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