Solve sqrt{4x^2-9x+5}-sqrt{2x^2-x-1}=sqrt{x^2-1}

$$\sqrt { 4 x ^ { 2 } - 9 x + 5 } - \sqrt { 2 x ^ { 2 } - x - 1 } = \sqrt { x ^ { 2 } - 1 }$$

Solve for x

$$x=1$$

$$x=-5$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{4x^{2}-9x+5}-\sqrt{2x^{2}-x-1}\right)^{2}=\left(\sqrt{x^{2}-1}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{4x^{2}-9x+5}-\sqrt{2x^{2}-x-1}\right)^{2}.

$$\left(\sqrt{4x^{2}-9x+5}\right)^{2}-2\sqrt{4x^{2}-9x+5}\sqrt{2x^{2}-x-1}+\left(\sqrt{2x^{2}-x-1}\right)^{2}=\left(\sqrt{x^{2}-1}\right)^{2}$$

Calculate \sqrt{4x^{2}-9x+5} to the power of 2 and get 4x^{2}-9x+5.

$$4x^{2}-9x+5-2\sqrt{4x^{2}-9x+5}\sqrt{2x^{2}-x-1}+\left(\sqrt{2x^{2}-x-1}\right)^{2}=\left(\sqrt{x^{2}-1}\right)^{2}$$

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

$$4x^{2}-9x+5-2\sqrt{4x^{2}-9x+5}\sqrt{2x^{2}-x-1}+2x^{2}-x-1=\left(\sqrt{x^{2}-1}\right)^{2}$$

Combine 4x^{2} and 2x^{2} to get 6x^{2}.

$$6x^{2}-9x+5-2\sqrt{4x^{2}-9x+5}\sqrt{2x^{2}-x-1}-x-1=\left(\sqrt{x^{2}-1}\right)^{2}$$

Combine -9x and -x to get -10x.

$$6x^{2}-10x+5-2\sqrt{4x^{2}-9x+5}\sqrt{2x^{2}-x-1}-1=\left(\sqrt{x^{2}-1}\right)^{2}$$

Subtract 1 from 5 to get 4.

$$6x^{2}-10x+4-2\sqrt{4x^{2}-9x+5}\sqrt{2x^{2}-x-1}=\left(\sqrt{x^{2}-1}\right)^{2}$$

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

$$6x^{2}-10x+4-2\sqrt{4x^{2}-9x+5}\sqrt{2x^{2}-x-1}=x^{2}-1$$

Subtract 6x^{2}-10x+4 from both sides of the equation.

$$-2\sqrt{4x^{2}-9x+5}\sqrt{2x^{2}-x-1}=x^{2}-1-\left(6x^{2}-10x+4\right)$$

To find the opposite of 6x^{2}-10x+4, find the opposite of each term.

$$-2\sqrt{4x^{2}-9x+5}\sqrt{2x^{2}-x-1}=x^{2}-1-6x^{2}+10x-4$$

Combine x^{2} and -6x^{2} to get -5x^{2}.

$$-2\sqrt{4x^{2}-9x+5}\sqrt{2x^{2}-x-1}=-5x^{2}-1+10x-4$$

Subtract 4 from -1 to get -5.

$$-2\sqrt{4x^{2}-9x+5}\sqrt{2x^{2}-x-1}=-5x^{2}-5+10x$$

Square both sides of the equation.

$$\left(-2\sqrt{4x^{2}-9x+5}\sqrt{2x^{2}-x-1}\right)^{2}=\left(-5x^{2}-5+10x\right)^{2}$$

Expand \left(-2\sqrt{4x^{2}-9x+5}\sqrt{2x^{2}-x-1}\right)^{2}.

$$\left(-2\right)^{2}\left(\sqrt{4x^{2}-9x+5}\right)^{2}\left(\sqrt{2x^{2}-x-1}\right)^{2}=\left(-5x^{2}-5+10x\right)^{2}$$

Calculate -2 to the power of 2 and get 4.

$$4\left(\sqrt{4x^{2}-9x+5}\right)^{2}\left(\sqrt{2x^{2}-x-1}\right)^{2}=\left(-5x^{2}-5+10x\right)^{2}$$

Calculate \sqrt{4x^{2}-9x+5} to the power of 2 and get 4x^{2}-9x+5.

$$4\left(4x^{2}-9x+5\right)\left(\sqrt{2x^{2}-x-1}\right)^{2}=\left(-5x^{2}-5+10x\right)^{2}$$

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

$$4\left(4x^{2}-9x+5\right)\left(2x^{2}-x-1\right)=\left(-5x^{2}-5+10x\right)^{2}$$

Use the distributive property to multiply 4 by 4x^{2}-9x+5.

$$\left(16x^{2}-36x+20\right)\left(2x^{2}-x-1\right)=\left(-5x^{2}-5+10x\right)^{2}$$

Use the distributive property to multiply 16x^{2}-36x+20 by 2x^{2}-x-1 and combine like terms.

$$32x^{4}-88x^{3}+60x^{2}+16x-20=\left(-5x^{2}-5+10x\right)^{2}$$

Square -5x^{2}-5+10x.

$$32x^{4}-88x^{3}+60x^{2}+16x-20=25x^{4}-100x^{3}+150x^{2}-100x+25$$

Subtract 25x^{4} from both sides.

$$32x^{4}-88x^{3}+60x^{2}+16x-20-25x^{4}=-100x^{3}+150x^{2}-100x+25$$

Combine 32x^{4} and -25x^{4} to get 7x^{4}.

$$7x^{4}-88x^{3}+60x^{2}+16x-20=-100x^{3}+150x^{2}-100x+25$$

Add 100x^{3} to both sides.

$$7x^{4}-88x^{3}+60x^{2}+16x-20+100x^{3}=150x^{2}-100x+25$$

Combine -88x^{3} and 100x^{3} to get 12x^{3}.

$$7x^{4}+12x^{3}+60x^{2}+16x-20=150x^{2}-100x+25$$

Subtract 150x^{2} from both sides.

$$7x^{4}+12x^{3}+60x^{2}+16x-20-150x^{2}=-100x+25$$

Combine 60x^{2} and -150x^{2} to get -90x^{2}.

$$7x^{4}+12x^{3}-90x^{2}+16x-20=-100x+25$$

Add 100x to both sides.

$$7x^{4}+12x^{3}-90x^{2}+16x-20+100x=25$$

Combine 16x and 100x to get 116x.

$$7x^{4}+12x^{3}-90x^{2}+116x-20=25$$

Subtract 25 from both sides.

$$7x^{4}+12x^{3}-90x^{2}+116x-20-25=0$$

Subtract 25 from -20 to get -45.

$$7x^{4}+12x^{3}-90x^{2}+116x-45=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -45 and q divides the leading coefficient 7. List all candidates \frac{p}{q}.

$$±\frac{45}{7},±45,±\frac{15}{7},±15,±\frac{9}{7},±9,±\frac{5}{7},±5,±\frac{3}{7},±3,±\frac{1}{7},±1$$

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 7x^{4}+12x^{3}-90x^{2}+116x-45 by x-1 to get 7x^{3}+19x^{2}-71x+45. Solve the equation where the result equals to 0.

$$7x^{3}+19x^{2}-71x+45=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 45 and q divides the leading coefficient 7. List all candidates \frac{p}{q}.

$$±\frac{45}{7},±45,±\frac{15}{7},±15,±\frac{9}{7},±9,±\frac{5}{7},±5,±\frac{3}{7},±3,±\frac{1}{7},±1$$

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 7x^{3}+19x^{2}-71x+45 by x-1 to get 7x^{2}+26x-45. Solve the equation where the result equals to 0.

$$7x^{2}+26x-45=0$$

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 7 for a, 26 for b, and -45 for c in the quadratic formula.

$$x=\frac{-26±\sqrt{26^{2}-4\times 7\left(-45\right)}}{2\times 7}$$

Do the calculations.

$$x=\frac{-26±44}{14}$$

Solve the equation 7x^{2}+26x-45=0 when ± is plus and when ± is minus.

$$x=-5 x=\frac{9}{7}$$

List all found solutions.

$$x=1 x=-5 x=\frac{9}{7}$$

Substitute 1 for x in the equation \sqrt{4x^{2}-9x+5}-\sqrt{2x^{2}-x-1}=\sqrt{x^{2}-1}.

$$\sqrt{4\times 1^{2}-9+5}-\sqrt{2\times 1^{2}-1-1}=\sqrt{1^{2}-1}$$

Simplify. The value x=1 satisfies the equation.

$$0=0$$

Substitute -5 for x in the equation \sqrt{4x^{2}-9x+5}-\sqrt{2x^{2}-x-1}=\sqrt{x^{2}-1}.

$$\sqrt{4\left(-5\right)^{2}-9\left(-5\right)+5}-\sqrt{2\left(-5\right)^{2}-\left(-5\right)-1}=\sqrt{\left(-5\right)^{2}-1}$$

Simplify. The value x=-5 satisfies the equation.

$$2\times 6^{\frac{1}{2}}=2\times 6^{\frac{1}{2}}$$

Substitute \frac{9}{7} for x in the equation \sqrt{4x^{2}-9x+5}-\sqrt{2x^{2}-x-1}=\sqrt{x^{2}-1}.

$$\sqrt{4\times \left(\frac{9}{7}\right)^{2}-9\times \frac{9}{7}+5}-\sqrt{2\times \left(\frac{9}{7}\right)^{2}-\frac{9}{7}-1}=\sqrt{\left(\frac{9}{7}\right)^{2}-1}$$

Simplify. The value x=\frac{9}{7} does not satisfy the equation because the left and the right hand side have opposite signs.

$$-\frac{4}{7}\times 2^{\frac{1}{2}}=\frac{4}{7}\times 2^{\frac{1}{2}}$$

Substitute 1 for x in the equation \sqrt{4x^{2}-9x+5}-\sqrt{2x^{2}-x-1}=\sqrt{x^{2}-1}.

$$\sqrt{4\times 1^{2}-9+5}-\sqrt{2\times 1^{2}-1-1}=\sqrt{1^{2}-1}$$

Simplify. The value x=1 satisfies the equation.

$$0=0$$

Substitute -5 for x in the equation \sqrt{4x^{2}-9x+5}-\sqrt{2x^{2}-x-1}=\sqrt{x^{2}-1}.

$$\sqrt{4\left(-5\right)^{2}-9\left(-5\right)+5}-\sqrt{2\left(-5\right)^{2}-\left(-5\right)-1}=\sqrt{\left(-5\right)^{2}-1}$$

Simplify. The value x=-5 satisfies the equation.

$$2\times 6^{\frac{1}{2}}=2\times 6^{\frac{1}{2}}$$

List all solutions of \sqrt{4x^{2}-9x+5}-\sqrt{2x^{2}-x-1}=\sqrt{x^{2}-1}.

$$x=1 x=-5$$