Solve sqrt{x^2+x-2}+sqrt{x^2-4x+3}=sqrt{x^2-1}

$$\sqrt { x ^ { 2 } + x - 2 } + \sqrt { x ^ { 2 } - 4 x + 3 } = \sqrt { x ^ { 2 } - 1 }$$

Solve for x

$$x=1$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{x^{2}+x-2}+\sqrt{x^{2}-4x+3}\right)^{2}=\left(\sqrt{x^{2}-1}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{x^{2}+x-2}+\sqrt{x^{2}-4x+3}\right)^{2}.

$$\left(\sqrt{x^{2}+x-2}\right)^{2}+2\sqrt{x^{2}+x-2}\sqrt{x^{2}-4x+3}+\left(\sqrt{x^{2}-4x+3}\right)^{2}=\left(\sqrt{x^{2}-1}\right)^{2}$$

Calculate \sqrt{x^{2}+x-2} to the power of 2 and get x^{2}+x-2.

$$x^{2}+x-2+2\sqrt{x^{2}+x-2}\sqrt{x^{2}-4x+3}+\left(\sqrt{x^{2}-4x+3}\right)^{2}=\left(\sqrt{x^{2}-1}\right)^{2}$$

Calculate \sqrt{x^{2}-4x+3} to the power of 2 and get x^{2}-4x+3.

$$x^{2}+x-2+2\sqrt{x^{2}+x-2}\sqrt{x^{2}-4x+3}+x^{2}-4x+3=\left(\sqrt{x^{2}-1}\right)^{2}$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}+x-2+2\sqrt{x^{2}+x-2}\sqrt{x^{2}-4x+3}-4x+3=\left(\sqrt{x^{2}-1}\right)^{2}$$

Combine x and -4x to get -3x.

$$2x^{2}-3x-2+2\sqrt{x^{2}+x-2}\sqrt{x^{2}-4x+3}+3=\left(\sqrt{x^{2}-1}\right)^{2}$$

Add -2 and 3 to get 1.

$$2x^{2}-3x+1+2\sqrt{x^{2}+x-2}\sqrt{x^{2}-4x+3}=\left(\sqrt{x^{2}-1}\right)^{2}$$

Calculate \sqrt{x^{2}-1} to the power of 2 and get x^{2}-1.

$$2x^{2}-3x+1+2\sqrt{x^{2}+x-2}\sqrt{x^{2}-4x+3}=x^{2}-1$$

Subtract 2x^{2}-3x+1 from both sides of the equation.

$$2\sqrt{x^{2}+x-2}\sqrt{x^{2}-4x+3}=x^{2}-1-\left(2x^{2}-3x+1\right)$$

To find the opposite of 2x^{2}-3x+1, find the opposite of each term.

$$2\sqrt{x^{2}+x-2}\sqrt{x^{2}-4x+3}=x^{2}-1-2x^{2}+3x-1$$

Combine x^{2} and -2x^{2} to get -x^{2}.

$$2\sqrt{x^{2}+x-2}\sqrt{x^{2}-4x+3}=-x^{2}-1+3x-1$$

Subtract 1 from -1 to get -2.

$$2\sqrt{x^{2}+x-2}\sqrt{x^{2}-4x+3}=-x^{2}-2+3x$$

Square both sides of the equation.

$$\left(2\sqrt{x^{2}+x-2}\sqrt{x^{2}-4x+3}\right)^{2}=\left(-x^{2}-2+3x\right)^{2}$$

Expand \left(2\sqrt{x^{2}+x-2}\sqrt{x^{2}-4x+3}\right)^{2}.

$$2^{2}\left(\sqrt{x^{2}+x-2}\right)^{2}\left(\sqrt{x^{2}-4x+3}\right)^{2}=\left(-x^{2}-2+3x\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4\left(\sqrt{x^{2}+x-2}\right)^{2}\left(\sqrt{x^{2}-4x+3}\right)^{2}=\left(-x^{2}-2+3x\right)^{2}$$

Calculate \sqrt{x^{2}+x-2} to the power of 2 and get x^{2}+x-2.

$$4\left(x^{2}+x-2\right)\left(\sqrt{x^{2}-4x+3}\right)^{2}=\left(-x^{2}-2+3x\right)^{2}$$

Calculate \sqrt{x^{2}-4x+3} to the power of 2 and get x^{2}-4x+3.

$$4\left(x^{2}+x-2\right)\left(x^{2}-4x+3\right)=\left(-x^{2}-2+3x\right)^{2}$$

Use the distributive property to multiply 4 by x^{2}+x-2.

$$\left(4x^{2}+4x-8\right)\left(x^{2}-4x+3\right)=\left(-x^{2}-2+3x\right)^{2}$$

Use the distributive property to multiply 4x^{2}+4x-8 by x^{2}-4x+3 and combine like terms.

$$4x^{4}-12x^{3}-12x^{2}+44x-24=\left(-x^{2}-2+3x\right)^{2}$$

Square -x^{2}-2+3x.

$$4x^{4}-12x^{3}-12x^{2}+44x-24=x^{4}-6x^{3}+13x^{2}-12x+4$$

Subtract x^{4} from both sides.

$$4x^{4}-12x^{3}-12x^{2}+44x-24-x^{4}=-6x^{3}+13x^{2}-12x+4$$

Combine 4x^{4} and -x^{4} to get 3x^{4}.

$$3x^{4}-12x^{3}-12x^{2}+44x-24=-6x^{3}+13x^{2}-12x+4$$

Add 6x^{3} to both sides.

$$3x^{4}-12x^{3}-12x^{2}+44x-24+6x^{3}=13x^{2}-12x+4$$

Combine -12x^{3} and 6x^{3} to get -6x^{3}.

$$3x^{4}-6x^{3}-12x^{2}+44x-24=13x^{2}-12x+4$$

Subtract 13x^{2} from both sides.

$$3x^{4}-6x^{3}-12x^{2}+44x-24-13x^{2}=-12x+4$$

Combine -12x^{2} and -13x^{2} to get -25x^{2}.

$$3x^{4}-6x^{3}-25x^{2}+44x-24=-12x+4$$

Add 12x to both sides.

$$3x^{4}-6x^{3}-25x^{2}+44x-24+12x=4$$

Combine 44x and 12x to get 56x.

$$3x^{4}-6x^{3}-25x^{2}+56x-24=4$$

Subtract 4 from both sides.

$$3x^{4}-6x^{3}-25x^{2}+56x-24-4=0$$

Subtract 4 from -24 to get -28.

$$3x^{4}-6x^{3}-25x^{2}+56x-28=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -28 and q divides the leading coefficient 3. List all candidates \frac{p}{q}.

$$±\frac{28}{3},±28,±\frac{14}{3},±14,±\frac{7}{3},±7,±\frac{4}{3},±4,±\frac{2}{3},±2,±\frac{1}{3},±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 3x^{4}-6x^{3}-25x^{2}+56x-28 by x-1 to get 3x^{3}-3x^{2}-28x+28. Solve the equation where the result equals to 0.

$$3x^{3}-3x^{2}-28x+28=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 28 and q divides the leading coefficient 3. List all candidates \frac{p}{q}.

$$±\frac{28}{3},±28,±\frac{14}{3},±14,±\frac{7}{3},±7,±\frac{4}{3},±4,±\frac{2}{3},±2,±\frac{1}{3},±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 3x^{3}-3x^{2}-28x+28 by x-1 to get 3x^{2}-28. Solve the equation where the result equals to 0.

$$3x^{2}-28=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 3 for a, 0 for b, and -28 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 3\left(-28\right)}}{2\times 3}$$

Do the calculations.

$$x=\frac{0±4\sqrt{21}}{6}$$

Solve the equation 3x^{2}-28=0 when ± is plus and when ± is minus.

$$x=-\frac{2\sqrt{21}}{3} x=\frac{2\sqrt{21}}{3}$$

List all found solutions.

$$x=1 x=-\frac{2\sqrt{21}}{3} x=\frac{2\sqrt{21}}{3}$$

Substitute 1 for x in the equation \sqrt{x^{2}+x-2}+\sqrt{x^{2}-4x+3}=\sqrt{x^{2}-1}.

$$\sqrt{1^{2}+1-2}+\sqrt{1^{2}-4+3}=\sqrt{1^{2}-1}$$

Simplify. The value x=1 satisfies the equation.

$$0=0$$

Substitute -\frac{2\sqrt{21}}{3} for x in the equation \sqrt{x^{2}+x-2}+\sqrt{x^{2}-4x+3}=\sqrt{x^{2}-1}.

$$\sqrt{\left(-\frac{2\sqrt{21}}{3}\right)^{2}-\frac{2\sqrt{21}}{3}-2}+\sqrt{\left(-\frac{2\sqrt{21}}{3}\right)^{2}-4\left(-\frac{2\sqrt{21}}{3}\right)+3}=\sqrt{\left(-\frac{2\sqrt{21}}{3}\right)^{2}-1}$$

Simplify. The value x=-\frac{2\sqrt{21}}{3} does not satisfy the equation.

$$2\times 7^{\frac{1}{2}}+3^{\frac{1}{2}}=\frac{5}{3}\times 3^{\frac{1}{2}}$$

Substitute \frac{2\sqrt{21}}{3} for x in the equation \sqrt{x^{2}+x-2}+\sqrt{x^{2}-4x+3}=\sqrt{x^{2}-1}.

$$\sqrt{\left(\frac{2\sqrt{21}}{3}\right)^{2}+\frac{2\sqrt{21}}{3}-2}+\sqrt{\left(\frac{2\sqrt{21}}{3}\right)^{2}-4\times \frac{2\sqrt{21}}{3}+3}=\sqrt{\left(\frac{2\sqrt{21}}{3}\right)^{2}-1}$$

Simplify. The value x=\frac{2\sqrt{21}}{3} does not satisfy the equation.

$$2\times 7^{\frac{1}{2}}-3^{\frac{1}{2}}=\frac{5}{3}\times 3^{\frac{1}{2}}$$

Substitute 1 for x in the equation \sqrt{x^{2}+x-2}+\sqrt{x^{2}-4x+3}=\sqrt{x^{2}-1}.

$$\sqrt{1^{2}+1-2}+\sqrt{1^{2}-4+3}=\sqrt{1^{2}-1}$$

Simplify. The value x=1 satisfies the equation.

$$0=0$$

Equation \sqrt{x^{2}+x-2}+\sqrt{x^{2}-4x+3}=\sqrt{x^{2}-1} has a unique solution.

$$x=1$$