第 10 类 RD Sharma 解——第 8 章二次方程——练习 8.1
问题 1. 以下哪些是二次方程?
(i) x 2 + 6x – 4 = 0
解决方案:
As we know that, quadratic equation is in form of: ax2 + bx + c = 0
Here, given equation is x2 + 6x – 4 =0 in quadratic equation.
So, It is a quadratic equations.
(ii) √3x 2 − 2x + ½= 0
解决方案:
As we know that, quadratic equation is in form of: ax2 + bx + c = 0
Here, given equation is √3x2 − 2x + ½= 0 in quadratic equation.
So, It is a quadratic equation.
(iii) x 2 + 1/x 2 = 5
解决方案:
This equation can be written as: x4 – 5x2 + 1 =0
As we know that, quadratic equation is in form of: ax2 + bx + c = 0
Here, given equation is x4 – 5x2 + 1 =0 is 4 degree polynomial.
So, It is not quadratic equations.
(iv) x – 3/x = x 2
解决方案:
This equation can be written as: x3 – x2 – 3 = 0
As we know that, quadratic equation is in form of: ax2 + bx + c = 0
Here, given equation is x3 – x2 – 3 = 0 is 3 degree polynomial.
So, It is not quadratic equations.
(v) 2x 2 – √3x + 9 = 0
解决方案:
As we know that, quadratic equation is in form of: ax2 + bx + c = 0
Here, given equation is 2x2 – √3x + 9 = 0 in quadratic equation.
So, It is a quadratic equation.
(vi) x 2 – 2x -√x – 5 = 0
解决方案:
As we know that, quadratic equation is in form of: ax2 + bx + c = 0
Here, given equation is x2 – 2x -√x -5 = 0.
So, It is not quadratic equation.
(vii) 3x 2 - 5x + 9 = x 2 - 7x + 3
解决方案:
This equation can be written as: 2x2 + 2x + 6 = 0
As we know that, quadratic equation is in form of: ax2 + bx + c = 0
Here, given equation is 2x2 + 2x + 6 = 0 in quadratic equation.
So, It is a quadratic equation.
(八) x + 1/x = 1
解决方案:
This equation can be written as: x2 – x + 1 = 0
As we know that, quadratic equation is in form of: ax2 + bx + c = 0
Here, given equation is x2 – x + 1 = 0 in quadratic equation.
So, It is a quadratic equation.
(ix) x 2 – 3x = 0
解决方案:
As we know that, quadratic equation is in form of: ax2 + bx + c = 0
Here, given equation is x2 – 3x =0 in quadratic equation.
So, It is a quadratic equation.
(x) (x + 1/x) 2 = 3(x +1/x)
解决方案:
((x2 + 1)/x)2 = 3((x2 + 1)/2)
This equation can be written as: x5 – 3x4 + 2x3 – 3x2 + x =0
As we know that, quadratic equation is in form of: ax2 + bx + c = 0
Here, given equation is x5 – 3x4 + 2x3 – 3x2 + x = 0 is 5 degree polynomial.
So, It is not quadratic equations.
(xi) (2x 2 + 1)(3x 2 + 2) = 6(x 2 - 1)(x 2 - 2)
解决方案:
This equation can be written as:
6x2 + 4x + 3x + 2 = 6x2 -12x – 6x + 12
7x + 2 = -18x + 12
This equation can be written as: 25x – 10 = 0
As we know that , quadratic equation is in form of: ax2 + bx + c = 0
Here, given equation is 25x – 10 = 0 is 1 degree polynomial.
So, It is not quadratic equations.
(xii) x + 1/x = x 2 , x ≠ 0
解决方案:
x + 1/x = x2
On multiplying by x on both sides we have,
x2 + 1 = x3
This equation can be written as: x3 – x2 – 1 = 0
As we know that, quadratic equation is in form of: ax2 + bx + c = 0
Here, given equation x3 – x2 – 1 = 0 is = 0 is 3 degree polynomial.
So, It is not quadratic equations.
(xiii) 16x 2 – 3 = (2x + 5)(5x – 3)
解决方案:
16x2 – 3 = (2x + 5)(5x – 3)
16x2 – 3 = 10x2 – 6x + 25x – 15
This equation can be written as: 6x2 – 19x + 12 = 0
As we know that , quadratic equation is in form of: ax2 + bx + c = 0
Here, given equation is 6x2 – 19x + 12 = 0 is 2 degree polynomial.
So, It is quadratic equations.
(xiv) (x + 2) 3 = x 3 – 4
解决方案:
(x + 2)3 = x3 – 4
On expanding, we get
x3 + 6x2 + 8x + 8 = x3 – 4
6x2 + 8x + 12 = 0
This equation can be written as: 6x2 + 8x + 12 = 0
As we know that, quadratic equation is in form of: ax2 + bx + c = 0
Here, given equation is 6x2 – 19x + 12 = 0 is 2 degree polynomial.
So, It is quadratic equations.
(xv) x(x + 1) + 8 = (x + 2)(x – 2)
解决方案:
x(x + 1) + 8 = (x + 2)(x – 2)
x2 + x + 8 = x2 – 4
x + 12 = 0
This equation can be written as: x + 12 = 0
As we know that, quadratic equation is in form of: ax2 + bx + c = 0
Here, given equation is x + 12 = 0 is 0 degree polynomial.
So, It is not a quadratic equation.
问题 2. 在以下每一项中,确定给定值是否是给定方程的解:
(i) x 2 – 3x + 2 = 0, x = 2, x = – 1
解决方案:
Here we have check values to determine the solution the given equation:
LHS = x2– 3x + 2
Now we will put x = 2 in LHS, we get
(2)2 – 3(2) + 2
⇒ 4 – 6 + 2 = 0 = RHS
⇒ LHS = RHS
So, we can say that x = 2 is not a solution of the given equation.
Similarly,
Now, we will put x = -1 in LHS, we get
(-1)2 – 3(-1) + 2
1 + 3 + 2 = 6 ≠ RHS
LHS ≠ RHS
So, we can say that x = -1 is not a solution of the given equation.
(ii) x 2 + x + 1 = 0, x = 0, x = 1
解决方案:
Here we have check values to determine the solution the given equation:
LHS = x2 + x + 1
Now we will put x = 0 in LHS, we get
(0)2 + 0 + 1
⇒ 1 ≠ RHS
⇒ LHS ≠ RHS
So , we can say that x = 0 is not a solution of the given equation.
Similarly,
Now, we will put x = 1 in LHS, we get
(1)2 + 1 + 1
⇒ 3 ≠ RHS
⇒ LHS ≠ RHS
So, we can say that x = 1 is not a solution of the given equation.
(iii) x 2 - 3√3x + 6 = 0, x = √3 和 x = -2√3
解决方案:
Here we have check values to determine the solution the given equation:
LHS = x2 − 3√3x + 6
Now, we will put x = √3 in LHS, we get
(√3)2 − 3√3(√3) + 6
⇒ 3 – 9 + 6 = 0 = RHS
⇒ LHS = RHS
So , we can say that x = √3 is a solution of the given equation.
Similarly,
Now, we will put x = −2√3 in LHS, we get
(-2√3)2 − 3√3(-2√3) + 6
⇒ 12 + 18 + 6 = 36 ≠ RHS
⇒ LHS ≠ RHS
So, we can say that x = −2√3 is not a solution of the given equation.
(iv) x + 1/x = 13/6, x = 5/6, x = 4/3
解决方案:
Here we have check values to determine the solution the given equation:
LHS = x +1/ x
Now, we will put x = 5/6 in LHS, we get
(5/6) + 1/(5/6) = 5/6 + 6/5
⇒ 61/30 ≠ RHS
⇒ LHS ≠ RHS
So , we can say that x = 5/6 is not a solution of the given equation.
Similarly,
Now, we will put x = 4/3 in LHS, we get
(4/3) + 1/(4/3) = 4/3 + 3/4
⇒ 25/12 ≠ RHS
⇒ LHS ≠ RHS
So, we can say that x = 3/4 is not a solution of the given equation.
(v) 2x 2 – x + 9 = x 2 + 4x + 3, x = 2, x = 3
解决方案:
Here we have check values to determine the solution the given equation:
2x2– x + 9 = x2 + 4x + 3
⇒ x2 – 5x + 6 = 0
LHS = x2– 5x + 6
Now, we will put x = 2 in LHS, we get
(2)2 – 5(2) + 6
4 – 10 + 6 = 0 = RHS
⇒ LHS = RHS
So , we can say that x = 2 is a solution of the given equation.
Similarly,
Now, we will put x = 3 in LHS, we get
(3)2 – 5(3) + 6
⇒ 9 – 15 + 6 = 0 = RHS
⇒ LHS = RHS
So, we can say that x = 3 is a solution of the given equation.
(vi) x 2 – √2x – 4 = 0, x = -√2, x = -2√2
解决方案:
Here we have check values to determine the solution the given equation:
LHS = x2 – √2x – 4
Now, we will put x = -√2 in LHS, we get
(-√2)2 − √2(-√2) – 4
⇒ 4 + 2 – 4 = 2 ≠ RHS
⇒ LHS ≠ RHS
So , we can say that x = -√2 is a solution of the given equation.
Similarly,
Now, we will put x = −2√2 in LHS, we get
(-2√2)2 − √2(-2√2) – 4
⇒ 8 + 4 – 4 = 8 ≠ RHS
⇒ LHS ≠ RHS
So, we can say that x = −2√2 is not a solution of the given equation.
(vii) a 2 x 2 – 3abx + 2b 2 = 0, x = a/b, x = b/a
解决方案:
Here we have check values to determine the solution the given equation:
LHS = a2x2 – 3abx + 2b2
Now, we will put x = a/b in LHS, we get
a2(a/b)2 -3ab(a/b) + 2b2
⇒ a4/b2 – 3a2 + 2b2 ≠ RHS
⇒ LHS ≠ RHS
So, we can say that x = a/b is a solution of the given equation.
Similarly,
Now, we will put x = b/a in LHS, we get
a2(b/a)2 – 3ab(b/a) + 2b2
⇒ b2 – 3b2 + 2b2 = RHS
⇒ LHS = RHS
So, we can say that x = b/a is not a solution of the given equation.
问题 3:在以下每一项中,找出给定值是给定方程的解的 k 值。
(i) 7x 2 + kx – 3 = 0, x = 2/3
解决方案:
Here we have to find value of k. So, we will put value of x = 2/3 in the given equation.
Here, x = 2/3
7(2/3)2 + k(2/3) – 3 =0
28/9 + 2k/3 -3 =0
2k/3 = 3 – 28/9
2k/3 = (27 – 28)/9
2k/3 = -1/9
k = -1/6
On putting value of x = 2/3 , we will get k =-1/6.
(ii) x 2 – x(a + b) + k = 0, x = a
解决方案:
Here we have to find value of k. So, we will put value of x = a in the given equation.
Here, x = a
a2 – a2 – ab + k = 0
k = ab
On putting value of x = a, we get k = ab.
(iii) kx 2 + √2x – 4 = 0, x = √2
解决方案:
Here we have to find value of k. So, we will put value of x = √2 in the given equation.
Here, x = √2
k(√2)2 + (√2)2 – 4 = 0
2k + 2 – 4 = 0
2k – 2 = 0
k = 1
On putting value of x = √2 , we get k = 1.
(iv) x 2 + 3ax + k = 0, x = -a
解决方案:
Here we have to find value of k. So, we will put value of x = -a in the given equation.
Here, x = -a
(- a)2 + 3a(- a) + k = 0
a2 – 3a2 + k = 0
k = 2a2
On putting value of x = – a, we get k = 2a2.
问题 4. 给定检查 3 是否是方程的根
√(x 2 – 4x +3) + √(x 2 -9) = √(4x 2 – 14x + 16)
解决方案:
Here we have to check, 3 is root of equation or not.
So, we first put x = 3 in LHS side.
√((3)2 – 4(3) + 3) + √((3)2 – 9)
⇒ 0 + 0 = 0
Now , we will put value x = 3 in RHS side.
√(4(3)2 – 14(3) + 16)
⇒ √(52 – 42) = √10
Now we can say that, LHS ≠ RHS.
So, x = 3 is not the root of the equation.
问题 5. 如果 x = 2/3 和 x = -3 是方程 ax 2 + 7x + b = 0 的根,求 a 和 b 的值。
解决方案:
Here we have to find value of a and b, so we will put values x = 2/3 and x = -3 in the equation.
So first put x = 2/3
a(2/3)2 + 7(2/3) + b =0
4a/9 + 14/3 + b =0
4a + 42 + 3b = 0 — (i)
Now, x = -3.
a(-3)2 + 7(-3) + b = 0
9a – 21 + b = 0 –(ii)
Now solving these equations (i) & (ii).
After solving these equations, we get a = 3 and b = -6.