第 10 类 RD Sharma 解决方案 - 第 8 章二次方程 - 练习 8.6 |设置 1

问题 1. 确定下列二次方程的根的性质：

(i) 2x² – 3x + 5 = 0

(ii) 2x² – 6x + 3 = 0

(iii) 3/5 x² – 2/3 x + 1 = 0

(iv) 3x² – 4√3 x + 4 = 0

(v) 3x² – 2√6 x + 2 = 0

解决方案：

(i) 2x² – 3x + 5 = 0

Here a=2, b=-3, c=5

D=b²-4ac=(-3)²-4*2*5

=-9-40=-31

D<0

Roots are not real

(ii) 2x² – 6x + 3 = 0

Here a=2, b=-6, c=3

D=b²-4ac

=(-6)²-4*2*3=36-24=12

D>0

Roots are real and distinct

(iii) 3/5 x² – 2/3 x + 1 = 0

Here a=3/5, b=-2/3, c=1

Discriminant (D)=b²-4ac

$=(\frac{-2}{3})^2-4*\frac{3}{5}*1\\ =\frac{4}{9}-\frac{12}{5}=\frac{20-108}{45}=\frac{-88}{45}$

D<0

Roots are not real

(iv) 3x² – 4√3 x + 4 = 0

Here a=3, b=-4√3, c=4

D=b²-4ac

=(-4√3)2-4*3*4=48-48=0

D=0

Roots are real and equal

(v) 3x² – 2√6 x + 2 = 0

Here a=3, b=– 2√6, c=2

Discriminant (D)=b²-4ac

=(– 2√6)2-4*3*2=24-24=0

D=0

Roots are real and equal

问题 2. 在以下每个方程中求根为实数且相等的 k 值：

(i) kx² + 4x + 1 = 0

解决方案：

Here a=k, b=4, c=1

Discriminant(D)=b2-4ac

=(4)²-4*k*1

=16-4k

Roots are real and equal

D=0

16-4k=0⇒4k=16

k=16/4=4

Hence k=4

(ii) kx² – 2√5 x + 4 = 0

解决方案：

Here a=k, b=-2√5, c=4

Discriminant(D)=b²-4ac

=( – 2√5 )-4*k*4=20-16k

Roots are real and equal

D=0

20-16k=0⇒ 16k=20

k=20/16=5/4

Hence k=5/4

(iii) 3x² – 5x + 2k = 0

解决方案：

Here a=3, b=-5, c=2k

Discriminant (D)=b²-4ac

=(-5)²-4*3*2k

=25-24k

Roots are real and equal

D=0

25-24k=0⇒24k=25

k=25/24

(iv) 4x²+ kx + 9 = 0

解决方案：

Here a=4, b=k, c=9

Discriminant (D)=b²-4ac

=k²-4*4*9=k2-144

Roots are real and equal

k²-144=0⇒k2=144=(±12)²

(v) 2kx² – 40x + 25 = 0

解决方案：

Here a=2k, b=-40, c=25

Discriminant(D)=b²-4ac

=(-40)²-4*2k*25

=1600-200k

Roots are real and equal

D=0

1600-200k=0⇒200k=1600

k=1600/200=8

Hence k=8

(vi) 9x² – 24x + k = 0

解决方案：

Here a=9, b=-24, c=k

Discriminant(D)=b²-4ac

=(-24)²-4*9*k

=576-36k

Roots are real and equal

D=0

576-36k=0

36k=576⇒k=576/36=16

k=16

(vii) 4x² – 3kx +1 = 0

解决方案：

Here a=4, b=-3k, c=1

Discriminant (D)=b²-4ac

=(-3k)²-4*4*1

=9k2-16

Roots are real and equal

D=0

9k²-16=0⇒9k2=16

k²=16/9=$(±\frac{4}{3})^2\\ k=±\frac{4}{3}$

(viii) x² – 2 (5 + 2k) x + 3 (7 + 10k) = 0

解决方案：

Here a=1, b=-2(5+2k) and c=3(7+10k)

Discriminant (D)=b²-4ac

[-2(5+2k)]²-4*1*3(7+10k)

=4(25+4k²+20k)-12(7+10k)

=100+16k²+80k-84-120k

16k²-40k+16

Roots are real and equal

D=0

16k²-40k+16=0

2k²-5k+2=0

2k²-4k-k+2=0

2k(k-2)-1(k-2)=0

(k-2)(2k-1)=0

Either k-2=0, then k=2 or 2k-1=0, then 2k=1⇒k=1/2

Hence, k=2, 1/2

(ix) (3k + 1) x² + 2(k + 1) x + k = 0

解决方案：

Here a=3k+1, b=2(k+1), c=k

Discriminant(D)=b²-4ac

=[2(k+1)]²-4*(3k+1)*k

=4(k²+2k+1)-4k(3k+1)

=4k²+8k+4-12k2-4k

-8k²+4k+4

Roots are real and equal

D=0

-8k²+4k+4=0

2k²-k-1=0 (Dividing by -4)

2k²-2k+k-1= {Therefore -2=-2*1

-1=-2+1}

2k(k-1)+1(k-1)=0

(k-1)(2k+1)=0

Either k-1=0, then k=1 or 2k+1=0, then 2k=-1⇒k=-1/2

k=1,-1/2

(x) kx² + kx + 1 = – 4x² – x

解决方案：

kx² +4x²+kx+x+1=0

(k+4)x²+(k+1)x+1=0

Here a=k+4, b=k+1, c=1

Discriminant(D)=b²-4ac

=(k+1)²-4*(k+4)*1

=k²+2k+1-4k-16

=k²-2k-15

Roots are real and equal

D=0

k²-2k-15=0

k²-5k+3k-15=0 {Therefore -15=-5*3

-2=-5+3}

k(k-5)+3(k-5)=0

(k-5)(k+3)=0

Either k-5=0, then k=5

or k+3=0, then k=-3

Hence k=5,-3

(xi) (k + 1) x² + 2 (k + 3) x + (k + 8) = 0

解决方案：

Here a=k+1, b=2(k+3), c=k+8

Discriminant(D)=b²-4ac

=[2(k+3)]²-4(k+1)(k+8)

=4(k²+6k+9)-4)(k2+9k+8)

=4k²+24k+36-4k²-36k-32

=-12k+4

Roots are real and equal

D=0

-12k+4=0

12k=4⇒k=4/12=1/3

Hence k=1/3

(xii) x² – 2kx + 7k – 12 = 0

解决方案：

Here a=1, b=-2k, c=(7k-12)

Discriminant(D)=b²-4ac

=(-2k)²-4*1*(7k-12)

=4k²-4(7k-12)

=4k²-28k+48

Roots are real and equal

D=0

4k²-28k+48=0

k²-7k+12=0 (Dividing by 4 )

k²-3k-4k+12=0 {12=-3*(-4)

-7=-3-4}

k(k-3)-4(k-3)=0

(k-3)(k-4)=0

Either, k-3=0, then k=3

or k-4=0, then k=4

Therefore, k=3,4

(xiii) (k + 1) x² – 2 (3k + 1) x + 8k + 1 = 0

解决方案：

Here a=k+1, b=-2(3k+1), c=8k+1

Discriminant(D)=b²-4ac

=[-2(3k+1)²-4*(k+1)(8k+1)]

=4(9k²+6k+1)-4(8k2+9k+1)

=36k²+24k+4-32k²-36-4

=4k²-12k

Roots are real and equal

D=0

4k²-12k=0

k²-3k=0 ————–(Dividing by 4)

k(k-3)=0

Either k=0

or k-3=0, then k=3

k=3,0

(xiv) 5x² – 4x + 2 + k (4x² – 2x – 1) = 0

解决方案：

5x²-4x+2+4kx2-2kx-k=0

(5+4k)x²-(4+2k)x+(2-k)=0

Here a=5+4k, b=-(4+2k), c=2-k

Discriminant (D)=b²-4ac

=[-(4+2k)]²-4*(5+4k)(2-k)

=16+4k²+16k-4(10-5k+8k-4k²)

=16+4k²+16k-40+20k-32k+16k²

=20k²+4k-24

Roots are real and equal

D=0

20k²+4k-24=0

5k²+k-6=0 —–(Dividing by 4)

5k²+6k-5k-6=0

k(5k+6)-1(5k+6)=0

(5k+6)(k-1)=0

Either 5k+6=0, then 5k=-6⇒k=-6/5

or k-1=0, then k=1

k=1,$\frac{-6}{5}$

(xv) (4 – k) x² + (2k + 4) x (8k + 1) = 0

解决方案：

Here a=4-k, b=2k+4, c=8k+1

Discriminant (D)=b²-4ac

=(2k+4)²-4*(4-k)(8k+1)

=4k²+16k+16-4(32k+4-8k²-k)

=4k²+16k+16-128k-16+32k²+4k

=36k²-108k

Roots are real and equal

36k²-108k=0

k²-3k=0⇒k(k-3)=0

Either k=0

or k-3=0, then k=3

Hence k=0,3

(xvi) (2k + 1) x² + 2 (k + 3) x (k + 5) = 0

解决方案：

Here a=2k+1, b=2(k+3), c=k+5

Discriminant (D)=b²-4ac

=[2(k+3)]²-4(2k+1)(k+5)

=4(k²+6k+9)-4(2k²+10k+k+5)

=4k²+24k+36-8k²-40k-4k-20

=-4k²-20k+16

Roots are real and equal D=0

-4k²-20k+16=0

k²+5k-4=0 ———-(Dividing by -4)

Here a=1, b=5, c=-4

Discriminant (D)=b²-4ac

=(5)²-4*1*(-4)=25+16=41

$k=\frac{-b±\sqrt{b^2-4ac}}{2a}=\frac{-5±\sqrt{41}}{2*1}$

$=\frac{-5+\sqrt{41}}{2},\frac{-5-\sqrt{41}}{2}$

(xvii) 4x² – 2 (k + 1) x + (k + 4) = 0

解决方案：

Here a=4, b=-2(k+1), c=k+4

Discriminant (D)=b²-4ac

=[-2(k+1)]²-4*4*(k+4)

=4(k²+2k+1)-16(k+4)

=4k²+8k+4-16k-64

=4k²-8k-12

Roots are real and equal

D=0

4k²-8k-60=0

k²-2k-15=0 ————-(Dividing by 4)

k²-5k+3k-15=0

k(k-5)+3(k-5)=0

(k-5)(k+3)=0

Either k-5=0, then k=5

or k+3=0, then k=-3

k=5,-3

(xviii) 4x² (k + 1) x + (k + 1) = 0

解决方案：

Here a=4, b=-2(k+1), c=k+1

Discriminant (D)=b²-4ac

=[-2(k+1)]²-4*4*(k+1)

=4(k²+2k+1)-16(k+1)

=4k²+8k+4-16k-16

=4k²-8k-12

k²-2k-3=0 ————(Dividing by 4)

k²-3k²+k-3=0

k(k-3)+(k-3)=0

(k-3)(k+1)=0

Either (k-3)=0, then k=3

or (k+1)=0, then k=-1

k=-1,3

问题 3. 在下面，确定给定二次方程具有实根的 k 值的集合：

(i) 2x² + 3x + k = 0

解决方案：

Here a=2, b=3, c=k

Discriminant (D)=b²-4ac

=(3)²-4*2*k

=9-8k

The roots are real

D≥0

9-8k≥0⇒9≥8k⇒8k≤9

k≤9/8

(ii) 2x² + x + k = 0

解决方案：

Here a=2, b=1, c=k

Discriminant (D)=b²-4ac

=(1)²-4*2*k

=1-8k

The roots are real

D≥0

1-8k≥0⇒1≥8k

8k≤1

k≤1/8

(iii) 2x² – 5x – k = 0

解决方案：

Discriminant (D)=b²-4ac

=(-5)²-4*2*(-k)

=25+8k

Roots are real

D≥0

25+8k≥0

8k≥-25⇒≥-25/8

k≥-25/8

(iv) kx² + 6x + 1 = 0

解决方案：

Discriminant (D)=b²-4ac

=(6)²-4*k*1

=36-4k

Roots are real

D≥0⇒36-4k≥0

36≥4k⇒4k≤36

k≤36/4⇒k≤9

k≤9

(v) 3x² + 2x + k = 0

解决方案：

Here a=3, b=2, c=k

Discriminant (D)=b²-4ac

=(2)2-4*3*k=4-12k

Roots are real

D≥0⇒4-12k≥0

4≥12k⇒12k≤0

4≥12k⇒12k≤4

k≤$\frac{4}{12}$⇒k≤1/3

问题 4. 找出下列方程具有实根和相等根的 k 值：

(i) x²- 2(k + 1) x + k² = 0

解决方案：

Here a=1, b=2(k+1), c=k²

Discriminant (D)=b²-4ac

=[2(k+1)]²-4*1*k²

=4(k²+2k+1)-4k²

=4k²+8k+4-k²

=8k+4

Roots are real and equal

D=0

8k+4=0⇒8k=-4

k=-4/8=-1/2, Hence k=-1/2

(ii) k²x² – 2 (2k – 1) x + 4 = 0

解决方案：

Here, a=k2, b=-2(2k-1), c=4

Discriminant (D)=b²-4ac

=[-2(2k-1)]²-4*k²*4

=4(4k²-4k+1)-16k²

=16k²-16k+4-16k²=-16k+4

Roots are real and equal

D=0

-16k+4=0⇒-16k=-4

k=4/16=1/4

k=1/4

(iii) (k + 1) x² – 2(k – 1) x + 1 = 0

解决方案：

Here, a=k+1, b=-2(k-1) and c=1

Discriminant (D)=b²-4ac

=[-2(k-1)]²-4(k+1)*1

=4(k²-2k+1)-4(k+1)

=4k²-8k+4-4k-4=4k²-12k

Roots are real and equal

D=0

4k2-12k=0

k2-3k=0 ————-(Dividing by 4)

Either k=0

or k-3=0, then k=3

k=0,3

(iv) x² + k(2x + k – 1) + 2 = 0

解决方案：

Here, a=1, b=2k, c=(k²-k+2)

Discriminant (D)=b²-4ac

=(2k)²-4*1*(k²-k+2)

=4k²-4k²+4k-8

=4k-8

Roots are real and equal

D=0

4k-8=0⇒k=2

Hence, k=2

问题 5. 求下列方程有实根的 k 值

(i) 2x² + kx + 3 = 0

解决方案：

Here a = 2, b = k, c = 3

Roots are real and equal

D=0

k²-24=0⇒k²=24

k=±√24=±√4*6=±2√6

(ii) kx (x – 2) + 6 = 0

解决方案：

kx²-2kx+6=0

here, a=k, b=-2k, c=6

Discriminant (D)=b²-4ac=(-2k)²-4*k*6=4k²-24k

Roots are real and equal

D=0

4k²-24k=0⇒4k(k-6)=0

k(k-6)=0

Either k=0 or

k-6=0, then k=6

k=0,6

(iii) x² – 4kx + k = 0

解决方案：

Here, comparing with ax²+kx+c=0

a=1, b=-4k, c=k

Discriminant (D)=b²-4ac=(-4k)2-4*1*k=16k2-4k

Roots are real and equal

D=0

16k²-4k=0⇒4k(4k-1)=0

k(4k-1)=0

Either k=0

or 4k-1=0⇒4k=1

k=1/4, Hence k=0,1/4

(iv) kx(x – 2√5) + 10 = 0

解决方案：

Here a=k, b=-2√5k, c=10

D=b²-4ac

=(-2√5k)²-4(k)(10)=20k²-40k

Roots are equal D=0

20k²-40k=0

k-2=0 ———(Dividing by 20k)

k=2

(v) kx (x – 3) + 9 = 0

解决方案：

Here, a=k, b=-3k, c=9

D=b²-4ac

=(-3k)²-4(k)9

=9k²-36k

For roots to be real

D=0

9k²-36k=0

9k(k-4)=0

k-4=0⇒k=4

k=4

(vi) 4x² + kx + 3 = 0

解决方案：

Here, a=4, b=k, c=3

D=b²-4ac

=k2-4(4)(3)

=k2-48

For roots to be real

D=0

k2-48=0

k2=48

k=±√48=±$\sqrt{16*3}$

k=±$4\sqrt{3}$

问题 6. 求给定二次方程具有实数和不同根的 k 值：

(i) kx² + 2x + 1 = 0

解决方案：

Here, a=k, b=2, c=1

D=b²-4ac

=(2)²-4*k*1

=4-4k

Roots are real and distinct

D>0⇒4-4k>0

1-k>0⇒1>k

⇒k<1

Therefore, k<1

(ii) kx² + 6x + 1 = 0

解决方案：

Here, a=k, b=6, c=1

D=b²-4ac

=(6)2-4*k*1

=36-4k

Roots are real and distinct

D>0⇒36-4k>0

9-k>0⇒9>k

⇒k<9

Therefore, k<9

问题 7. 对于什么 k 值，(4 – k) x² + (2k + 4) x + (8k + 1) = 0，是一个完美的正方形。

解决方案：

(4 – k) x² + (2k + 4) x + (8k + 1) = 0

Here, a = 4 – k, b = 2k + 4, c = 8k + 1

=(2k+4)²-4*(4-k)(8k+1)

=4k²+16k+16-4(32k+4-8k²-k)

=4k²+16k+16-4(-8k²+31k+4)

=4k²+16k+16+32k²-124k-16

=36k²-108k

Therefore, the given quadratic equation is a perfect square

The roots are real and equal

D=0⇒36k²-108k=0

Either k=0

or k-3=0⇒k=3

k=0,3

问题 8. 求方程 x² + kx + 4 = 0 具有实根的 k 的最小正值。

解决方案：

x² + kx + 4 = 0

Here, a=1, b=k, c=4

Therefore, Discriminant(D)=b²-4ac

=(k)²=4*1*4

=k²-16

It has real roots

D≥0⇒k²-16≥0

⇒k2≥16⇒(k)²≥(±4)²

k≥4 or k≤-4

Least positive value of k=4

问题 9. 求二次方程 (3k + 1) x² + 2(k + 1) x + 1 = 0 具有相等根的 k 值。另外，找根。

解决方案：

(3k + 1) x² + 2(k + 1) x + 1 = 0

Here a=(3k+1), b=2(k+1) ,c=1

Now, b²-4ac=[2(k+1)]²-4*(3k+1)*1

=4(k²+2k+1)-4(3k+1)

=4k²+8k+4-12k-4

=4k²-4k

Roots are real and equal

b²-4ac=0

4k²-4k=0

k²-k=0 k(k-1)=0

Either k=0 or k-1=0, then k=1

k=0,1

(i) When k=0, then

(3*0*1)x²+2(0+1)x+1=0

x²+2x+1=0

(x+1)²=0

x+1=0

x=-1

When k=1, then

(3*1+1)x²+2(1+1)x+1=0

4x²+4x²+1=0

(2x+1)²=0

2x+1=0

2x=-1⇒x=-1/2

x=1, -1/2

问题 10. 求二次方程 (2p + 1) x² – (7p + 2) x + (7p – 3) = 0 具有相等根的 p 值。另外，找到这些根源。

解决方案：

Here, a=2-+1, b=-(7p+2), c=(7p-3)

D=0 [Equal roots]

As b²-4ac=0

[-(7p+)]²-4(2p+1)(7p-3)=0

(7p+2)²-4(14p²-6p+7p-3)=0

49p²+28p+4-56p²+24p-28p+12=0

-7p²+24p+16=0

7p²-24-16=0 ————-(Dividing both sides by -1)

7p(p-4)+4(p-4)=0

(p-4)(7p+4)=0

p-4=0 or 7p+4=0

p=4 or p=-4/7

Roots are x=-b/2a [As equal roots (given)]

$x=\frac{(7p+2)}{2(2p+1)}$

Where p=4,

$\frac{(28+2)}{2(8+1)}=\frac{30}{18}=\frac{5}{3}$

Equal roots is 5/3

When p=-4/7

$x=\frac{[7(\frac{-4}{7})+2]}{[2(\frac{-4}{7}+1)]}\\ =\frac{\frac{(-28+14)}{7}}{2\frac{-8+7}{7}}=\frac{-14}{-2}=7$

Equal roots are 5/3 and 7