12 multiple-choice questions, progressively harder.
For a quadratic ax2+bx+cax^2 + bx + cax2+bx+c with real roots, (r1−r2)2(r_1 - r_2)^2(r1−r2)2 equals which expression?
Solution
Correct answer: C
Start from (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 and substitute r1+r2=−bar_1 + r_2 = -\frac{b}{a}r1+r2=−ab and r1r2=car_1 r_2 = \frac{c}{a}r1r2=ac.
(r1−r2)2=b2a2−4ca=b2−4aca2=Δa2(r_1 - r_2)^2 = \frac{b^2}{a^2} - \frac{4c}{a} = \frac{b^2 - 4ac}{a^2} = \frac{\Delta}{a^2}(r1−r2)2=a2b2−a4c=a2b2−4ac=a2Δ
So the squared gap between the roots is the discriminant divided by a2a^2a2.
The roots of x2−6x+kx^2 - 6x + kx2−6x+k differ by 222. Find kkk.
Correct answer: A
The condition is (r1−r2)2=22=4(r_1 - r_2)^2 = 2^2 = 4(r1−r2)2=22=4. For this monic quadratic, (r1−r2)2=(r1+r2)2−4r1r2=36−4k(r_1 - r_2)^2 = (r_1 + r_2)^2 - 4 r_1 r_2 = 36 - 4k(r1−r2)2=(r1+r2)2−4r1r2=36−4k.
36−4k=4 ⇒ k=836 - 4k = 4 \;\Rightarrow\; k = 836−4k=4⇒k=8
The discriminant 36−4(8)=4≥036 - 4(8) = 4 \ge 036−4(8)=4≥0, so the roots are real.
For the roots of x2−4x+1x^2 - 4x + 1x2−4x+1, find r13+r23r_1^3 + r_2^3r13+r23.
Correct answer: D
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 444 and product 111.
r13+r23=43−3(1)(4)=64−12=52r_1^3 + r_2^3 = 4^3 - 3(1)(4) = 64 - 12 = 52r13+r23=43−3(1)(4)=64−12=52
The identity turns a power sum into the sum and product.
The quadratic x2+bx+18x^2 + bx + 18x2+bx+18 has one root that is twice the other, and both roots are positive. Find bbb.
Correct answer: B
Let the roots be rrr and 2r2r2r. Their product is the constant 181818.
r⋅2r=2r2=18 ⇒ r=3r \cdot 2r = 2r^2 = 18 \;\Rightarrow\; r = 3r⋅2r=2r2=18⇒r=3
The roots are 333 and 666, so their sum is 999. Since the sum is −b-b−b, we get b=−9b = -9b=−9.
For which value of mmm does x2−8x+mx^2 - 8x + mx2−8x+m have two equal real roots?
Equal roots require Δ=(−8)2−4m=0\Delta = (-8)^2 - 4m = 0Δ=(−8)2−4m=0.
64−4m=0 ⇒ m=1664 - 4m = 0 \;\Rightarrow\; m = 1664−4m=0⇒m=16
Then the quadratic is x2−8x+16=(x−4)2x^2 - 8x + 16 = (x - 4)^2x2−8x+16=(x−4)2.
The two roots of x2−7x+cx^2 - 7x + cx2−7x+c are reciprocals of each other. Find ccc.
If the roots are reciprocals, r1=1r2r_1 = \frac{1}{r_2}r1=r21, so their product is 111. The product equals ccc for this monic quadratic.
r1r2=c=1r_1 r_2 = c = 1r1r2=c=1
The discriminant 49−4=45≥049 - 4 = 45 \ge 049−4=45≥0, so the roots are real.
The roots of x2+bx+12x^2 + bx + 12x2+bx+12 differ by 111. Find all possible real values of bbb.
The condition (r1−r2)2=1(r_1 - r_2)^2 = 1(r1−r2)2=1 becomes (r1+r2)2−4r1r2=b2−4(12)=1(r_1 + r_2)^2 - 4 r_1 r_2 = b^2 - 4(12) = 1(r1+r2)2−4r1r2=b2−4(12)=1.
b2−48=1 ⇒ b2=49 ⇒ b=±7b^2 - 48 = 1 \;\Rightarrow\; b^2 = 49 \;\Rightarrow\; b = \pm 7b2−48=1⇒b2=49⇒b=±7
Either sign gives real roots that differ by 111.
The equation x2−10x+q=0x^2 - 10x + q = 0x2−10x+q=0 has real roots. What is the largest value qqq can take?
Real roots require Δ=(−10)2−4q≥0\Delta = (-10)^2 - 4q \ge 0Δ=(−10)2−4q≥0.
100−4q≥0 ⇒ q≤25100 - 4q \ge 0 \;\Rightarrow\; q \le 25100−4q≥0⇒q≤25
The largest value is q=25q = 25q=25, where the roots become equal.
A quadratic has roots with r1+r2=6r_1 + r_2 = 6r1+r2=6 and r1r2=4r_1 r_2 = 4r1r2=4. Find r12r2+r1r22r_1^2 r_2 + r_1 r_2^2r12r2+r1r22.
Factor out r1r2r_1 r_2r1r2: the expression is r1r2(r1+r2)r_1 r_2 (r_1 + r_2)r1r2(r1+r2).
r12r2+r1r22=r1r2(r1+r2)=4⋅6=24r_1^2 r_2 + r_1 r_2^2 = r_1 r_2 (r_1 + r_2) = 4 \cdot 6 = 24r12r2+r1r22=r1r2(r1+r2)=4⋅6=24
It reduces to the product times the sum.
Do two real numbers with sum 444 and product 555 exist?
Test the condition s2−4p≥0s^2 - 4p \ge 0s2−4p≥0 with s=4s = 4s=4 and p=5p = 5p=5.
42−4(5)=16−20=−4<04^2 - 4(5) = 16 - 20 = -4 < 042−4(5)=16−20=−4<0
The condition fails, so no real numbers have sum 444 and product 555.
For 5x2−4x+c5x^2 - 4x + c5x2−4x+c to have real roots, which condition must ccc satisfy?
Real roots require Δ=(−4)2−4(5)c≥0\Delta = (-4)^2 - 4(5)c \ge 0Δ=(−4)2−4(5)c≥0.
16−20c≥0 ⇒ c≤1620=4516 - 20c \ge 0 \;\Rightarrow\; c \le \frac{16}{20} = \frac{4}{5}16−20c≥0⇒c≤2016=54
So ccc must be at most 45\tfrac{4}{5}54.
For the roots of x2−6x+7x^2 - 6x + 7x2−6x+7, 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 666 and product 777, the numerator is 62−2(7)=226^2 - 2(7) = 2262−2(7)=22.
r1r2+r2r1=227\frac{r_1}{r_2} + \frac{r_2}{r_1} = \frac{22}{7}r2r1+r1r2=722
Both pieces reduce to the sum and product.
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