Answer to Question #152744 in Algebra for Hardy

Question

Solving quadric equations by factorization

Find the real solutions of the following equations
1.(x^2+2x-14)^5=1
2.(x^2+7x-11)^8=1

Answer 1

1.

$$(x^2+2x-14)^5=1, x\in\R$$

Let $x^2+2x-14=u, u\in\R$

$$u^5-1=u^5-u^4+u^4-u^3+u^3-u^2+u^2-u+u-1$$

$$=(u-1)(u^4+u^3+u^2+u+1)$$

$$=(u-1)(u^4+u^3+\dfrac{1}{4}u^2+\dfrac{3}{4}u^2+u+1)$$

$$=(u-1)(u^2(u+\dfrac{1}{2})^2+\dfrac{1}{2}u^2+\dfrac{1}{4}u^2+u+1)$$

$$=(u-1)(u^2(u+\dfrac{1}{2})^2+\dfrac{1}{2}u^2+(\dfrac{1}{2}u+1)^2)$$

$$u^2(u+\dfrac{1}{2})^2+\dfrac{1}{2}u^2+(\dfrac{1}{2}u+1)^2>0, u\not=0, u\not=-\dfrac{1}{2}$$

$$u^2(u+\dfrac{1}{2})^2+\dfrac{1}{2}u^2+(\dfrac{1}{2}u+1)^2=1>0, u=0$$

$$u^2(u+\dfrac{1}{2})^2+\dfrac{1}{2}u^2+(\dfrac{1}{2}u+1)^2=\dfrac{1}{8}>0, u=-\dfrac{1}{2}$$

Then

$$u^2(u+\dfrac{1}{2})^2+\dfrac{1}{2}u^2+(\dfrac{1}{2}u+1)^2>0, u\in\R$$

Hence

$$u-1=0$$

$$x^2+2x-14-1=0$$

$$x^2+2x-15=0$$

$$(x+5)(x-3)=0$$

$$x+5=0\ or\ x-3=0$$

$$x_1=-5, x_2=3$$

$\{-5, 3\}$

2.

$$(x^2+7x-11)^8=1, x\in\R$$

$$((x^2+7x-11)^4-1)((x^2+7x-11)^4+1)=0$$

Since $x\in \R,$ then $(x^2+7x-11)^4+1>0$

$$(x^2+7x-11)^4-1=0$$

$$((x^2+7x-11)^2-1)((x^2+7x-11)^2+1)=0$$

Since $x\in \R,$ then $(x^2+7x-11)^2+1>0$

$$(x^2+7x-11)^2-1=0$$

$$((x^2+7x-11)-1)((x^2+7x-11)+1)=0$$

$$(x^2+7x-12)(x^2+7x-10)=0$$

$$x^2+7x-12=0\ or\ x^2+7x-10=0$$

$$x^2+7x-12=0$$

$$x^2+2(\dfrac{7}{2})x+(\dfrac{7}{2})^2-\dfrac{97}{4}=0$$

$$(x+\dfrac{7}{2})^2-\dfrac{97}{4}=0$$

$$x_1=-\dfrac{7}{2}-\dfrac{\sqrt{97}}{2}, x_2=-\dfrac{7}{2}+\dfrac{\sqrt{97}}{2}$$

$$x^2+7x-10=0$$

$$x^2+2(\dfrac{7}{2})x+(\dfrac{7}{2})^2-\dfrac{89}{4}=0$$

$$(x+\dfrac{7}{2})^2-\dfrac{89}{4}=0$$

$$x_3=-\dfrac{7}{2}-\dfrac{\sqrt{89}}{2}, x_4=-\dfrac{7}{2}+\dfrac{\sqrt{89}}{2}$$

$\bigg\{-\dfrac{7+\sqrt{97}}{2},-\dfrac{7+\sqrt{89}}{2},\dfrac{-7+\sqrt{89}}{2},\dfrac{-7+\sqrt{97}}{2}\bigg\}$