10类RD Sharma解–第8章二次方程式–练习8.4

问题1：通过完成平方的方法找到以下二次方的根(如果存在)： $x^2-4\sqrt{2x}+6=0$ 。

解决方案：

Given: $x^2-4\sqrt{2x}+6=0$

We have to make the equation a perfect square.

=> $x^2-4\sqrt{2x} = -6$

=> $x^2 - 2\times2\sqrt{2x} + (2\sqrt{2})^2 = -6 + (2\sqrt{2})^2$

We know that:

=> (a−b)² = a²−2×a×b+b²

Thus, the equation can be written as:

=> $(x-2\sqrt{2})^2 = -6 + (2\sqrt{2})^2$

=> $(x-2\sqrt{2})^2 = -6 + (2)^2(\sqrt{2})^2$

=> $(x-2\sqrt{2})^2 = -6 + (4\times2)$

=> $(x-2\sqrt{2})^2 = -6 + 8$

=> $(x-2\sqrt{2})^2 = 2$

The RHS is positive, which implies that the roots exist.

=> $(x-2\sqrt{2}) = \pm \sqrt{2}$

=> x = $\sqrt{2}+2\sqrt{2}$ and x= $-\sqrt{2}+2\sqrt{2}$

=> x = $3\sqrt{2}$ and x = $\sqrt{2}$

问题2：通过完成平方的方法找到以下二次方的根(如果存在)：2x ² -7x + 3 = 0。

解决方案：

Given: 2x²-7x+3 = 0

We have to make the equation a perfect square.

=> 2x²-7x+3 = 0

=> $x^2 - \dfrac{7}{2}x+\dfrac{3}{2}=0$

=> $x^2 -2\times \dfrac{7}{4}\times x+(\dfrac{7}{4})^2 - (\dfrac{7}{4})^2+\dfrac{3}{2} = 0$

We know that:

=> (a−b)²=a²−2×a×b+b²

Thus, the equation can be written as:

=> $(x-\dfrac{7}{4})^2 -\dfrac{49}{16} + \dfrac{3}{2}=0$

=> $(x-\dfrac{7}{4})^2 - \dfrac{25}{16} = 0$

=> $(x-\dfrac{7}{4})^2 = \dfrac{25}{16}$

The RHS is positive, which implies that the roots exist.

=> $(x-\dfrac{7}{4})^2 = \pm \dfrac{5}{4}$

=> $x = \dfrac{7}{4}+\dfrac{5}{4}$ and $x= \dfrac{7}{4}-\dfrac{5}{4}$

=> $x = \dfrac{12}{4}$ and $x = \dfrac{2}{4}$

=> x = 3 and $x = \dfrac{1}{2}$

问题3：通过完成平方的方法找到以下二次方的根(如果存在)：3x ² + 11x + 10 = 0。

解决方案：

Given: 3x²+11x+10 = 0

We have to make the equation a perfect square.

=> 3x²+11x+10 = 0

=> $x^2 + \dfrac{11}{3} x +\dfrac{10}{3}=0$

=> $x^2 + 2(\dfrac{11}{6})x+(\dfrac{11}{6})^2-(\dfrac{11}{6})^2+\dfrac{10}{3}=0$

We know that:

=> (a−b)²=a²−2×a×b+b²

Thus the equation can be written as:

=> $(x+\dfrac{11}{6})^2+\dfrac{10}{3}-\dfrac{121}{36}=0$

=> $(x+\dfrac{11}{6})^2 = \dfrac{1}{36}$

The RHS is positive, which implies that the roots exist.

=> $(x+\dfrac{11}{6})= \pm \dfrac{1}{6}$

=> $x = -\dfrac{11}{6}+\dfrac{1}{6}$ and $x= -\dfrac{11}{6}-\dfrac{1}{6}$

=> $x = -\dfrac{10}{6}$ and $x = -\dfrac{12}{6}$

=> $x = -\dfrac{5}{3}$ and x = -2

问题4：通过完成平方的方法找到以下二次方的根(如果存在)：2x ² + x-4 = 0。

解决方案：

Given: 2x²+x-4 =0

We have to make the equation a perfect square.

=> 2x²+x-4 =0

=> $x^2 - \dfrac{x}{2}-2=0$

=> $x^2 + 2\times \dfrac{1}{4}\times x+(\dfrac{1}{4})^2-(\dfrac{1}{4})^2-2=0$

We know that:

=> (a−b)²=a²−2×a×b+b²

Thus the equation can be written as:

=> $(x+\dfrac{1}{4})^2 = \dfrac{33}{16}$

The RHS is positive, which implies that the roots exist.

=>$(x+\dfrac{1}{4})= \pm \dfrac{\sqrt{33}}{4}$

=> $x = \dfrac{\sqrt{33}-1}{4}$ and $x= \dfrac{-\sqrt{33}-1}{4}$

问题5：通过完成平方的方法找到以下二次方的根(如果存在)：2x ² + x + 4 = 0。

解决方案：

Given: 2x²+x+4 =0

We have to make the equation a perfect square.

=> 2x²+x+4 =0

=> $x^2 + \dfrac{x}{2}+2=0$

=> $x^2 + 2\times \dfrac{1}{4}\times x+(\dfrac{1}{4})^2-(\dfrac{1}{4})^2+2=0$

We know that:

=> (a−b)²=a²−2×a×b+b²

Thus the equation can be written as:

=> $(x+\dfrac{1}{4})^2 = -\dfrac{31}{16}$

=> The RHS is negative, which implies that the roots are not real.

问题6：通过完成平方的方法找到以下二次方的根(如果存在)：4x ² +4√3+ 3 = 0。

解决方案：

Given: 4x²+4√3​+3=0

We have to make the equation a perfect square.

=> 4x²+4√3​+3=0

=> $x^2 + \sqrt{3}x+\dfrac{3}{4} =0$

=> $x^2 + 2\times x \times \dfrac{\sqrt{3}}{2} + (\dfrac{\sqrt{3}}{2})^2 - (\dfrac{\sqrt{3}}{2})^2+ \dfrac{3}{4}=0$

We know that,

=> (a−b)²=a²−2×a×b+b²

Thus the equation can be written as:

=> $(x+\dfrac{\sqrt{3}}{2})^2 = -\dfrac{3}{4}+(\dfrac{\sqrt{3}}{2})^2$

=> $(x+\dfrac{\sqrt{3}}{2})^2 = -\dfrac{3}{4}+\dfrac{3}{4}$

=> $(x+\dfrac{\sqrt{3}}{2})^2 = 0$

The RHS is zero, which implies that the roots exist and are equal.

=> $x = -\dfrac{\sqrt{3}}{2}$

问题7：通过完成正方形的方法找到以下二次方(如果存在)的根： $\sqrt{2}x^2 -3x-2\sqrt{2}=0$ 。

解决方案：

Given: $\sqrt{2}x^2 -3x-2\sqrt{2}=0$

We have to make the equation a perfect square.

=> $\sqrt{2}x^2 -3x-2\sqrt{2}=0$

=> $x^2 -\dfrac{3}{\sqrt{2}}x-2 =0$

=> $x^2 - 2\times x \times \dfrac{3}{2\sqrt{2}} + (\dfrac{3}{2\sqrt{2}})^2 - (\dfrac{3}{2\sqrt{2}})^2-2=0$

We know that,

=> (a−b)²=a²−2×a×b+b²

Thus the equation can be written as:

=> $(x-\dfrac{3}{2\sqrt{2}})^2 = 2 + (\dfrac{3}{2\sqrt{2}})^2$

=> $(x-\dfrac{3}{2\sqrt{2}})^2 = 2+ \dfrac{9}{8}$

=> $(x-\dfrac{3}{2\sqrt{2}})^2 = \dfrac{25}{8}$

The RHS is positive, which implies that the roots exist.

=> $(x-\dfrac{3}{2\sqrt{2}}) = \dfrac{5}{2\sqrt{2}}$

=> $x = \dfrac{5+3}{2\sqrt{2}}$ and $x = \dfrac{-5+3}{2\sqrt{2}}$

=> $x = 2\sqrt{2}$ and $x = \dfrac{-1}{\sqrt{2}}$

问题8：通过完成平方的方法找到以下二次方的根(如果存在)： $\sqrt{3}x^2+10x+7\sqrt{3}=0$ 。

解决方案：

Given: $\sqrt{3}x^2+10x+7\sqrt{3}=0$

We have to make the equation a perfect square.

=> $\sqrt{3}x^2+10x+7\sqrt{3}=0$

=> $x^2 +\dfrac{10}{\sqrt{3}}x+7 =0$

=> $x^2 +2\times x \times \dfrac{5}{\sqrt{3}} + (\dfrac{5}{\sqrt{3}})^2 - (\dfrac{5}{\sqrt{3}})^2+7=0$

We know that,

=> (a−b)²=a²−2×a×b+b²

Thus the equation can be written as:

=> $(x+\dfrac{5}{\sqrt{3}})^2 = -7 + (\dfrac{5}{\sqrt{3}})^2$

=> $(x+\dfrac{5}{\sqrt{3}})^2 = -7+ \dfrac{25}{3}$

=> $(x+\dfrac{5}{\sqrt{3}})^2 = \dfrac{4}{3}$

The RHS is positive, which implies that the roots exist.

=> $(x+\dfrac{5}{\sqrt{3}}) = \pm \dfrac{2}{\sqrt{3}}$

=> $x = \dfrac{-5+2}{\sqrt{3}}$ and $x = \dfrac{-5-2}{\sqrt{3}}$

=> $x = -\sqrt{3}$ and $x = \dfrac{-7}{\sqrt{3}}$

问题9：通过完成平方的方法找到以下二次方的根(如果存在)： $x^2-(\sqrt{2}+1)x+\sqrt{2}=0$ 。

解决方案：

Given: $x^2-(\sqrt{2}+1)x+\sqrt{2}=0$

We have to make the equation a perfect square.

=> $x^2-(\sqrt{2}+1)x+\sqrt{2}=0$

=> $x^2-(\sqrt{2}+1)x+(\dfrac{\sqrt{2}+1}{2})^2 - (\dfrac{\sqrt{2}+1}{2})^2+\sqrt{2}=0$

We know that,

=> (a−b)²=a²−2×a×b+b²

Thus the equation can be written as:

=> $(x-(\dfrac{\sqrt{2}+1}{2}))^2 = \dfrac{2+1+2\sqrt{2}}{4}-\sqrt{2}$

=> $(x-(\dfrac{\sqrt{2}+1}{2}))^2 = \dfrac{2+1-2\sqrt{2}}{4}$

=> $(x-(\dfrac{\sqrt{2}+1}{2}))^2 = (\dfrac{\sqrt{2}-1}{2})^2$

The RHS is positive, which implies that the roots exist.

=> $(x-(\dfrac{\sqrt{2}+1}{2})) = \pm (\dfrac{\sqrt{2}-1}{2})$

=> $x = \dfrac{\sqrt{2}-1+\sqrt{2}+1}{2}$ and $x = \dfrac{-\sqrt{2}+1+\sqrt{2}+1}{2}$

=> x = √2 and x = 1

问题10：通过完成平方的方法找到以下二次方程式的根(如果存在)：x ² -4ax + 4a ² -b ² = 0。

解决方案：

Given: x²-4ax+4a²-b²=0

We have to make the equation a perfect square.

=> x²-4ax+4a²-b²=0

=> x²−2×x×2a+(2a)²−b²=0

We know that,

=> (a−b)²=a²−2×a×b+b²

Thus the equation can be written as:

=> x²−2×2a×x+(2a)²=b²

=> (x-2a)² = b²

The RHS is positive, which implies that the roots exist.

=> (x-2a) = ±b

=> x= 2a+b and x = 2a-b