Solve 8(x)=sqrt{x^2-5x+6}

$$8 ( x ) = \sqrt { x ^ { 2 } - 5 x + 6 }$$

Solve for x

$$x=\frac{\sqrt{1537}-5}{126}\approx 0.271465012$$

Solution Steps

Square both sides of the equation.

$$\left(8x\right)^{2}=\left(\sqrt{x^{2}-5x+6}\right)^{2}$$

Expand \left(8x\right)^{2}.

$$8^{2}x^{2}=\left(\sqrt{x^{2}-5x+6}\right)^{2}$$

Calculate 8 to the power of 2 and get 64.

$$64x^{2}=\left(\sqrt{x^{2}-5x+6}\right)^{2}$$

Calculate \sqrt{x^{2}-5x+6} to the power of 2 and get x^{2}-5x+6.

$$64x^{2}=x^{2}-5x+6$$

Subtract x^{2} from both sides.

$$64x^{2}-x^{2}=-5x+6$$

Combine 64x^{2} and -x^{2} to get 63x^{2}.

$$63x^{2}=-5x+6$$

Add 5x to both sides.

$$63x^{2}+5x=6$$

Subtract 6 from both sides.

$$63x^{2}+5x-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 63 for a, 5 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\times 63\left(-6\right)}}{2\times 63}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\times 63\left(-6\right)}}{2\times 63}$$

Multiply -4 times 63.

$$x=\frac{-5±\sqrt{25-252\left(-6\right)}}{2\times 63}$$

Multiply -252 times -6.

$$x=\frac{-5±\sqrt{25+1512}}{2\times 63}$$

Add 25 to 1512.

$$x=\frac{-5±\sqrt{1537}}{2\times 63}$$

Multiply 2 times 63.

$$x=\frac{-5±\sqrt{1537}}{126}$$

Now solve the equation x=\frac{-5±\sqrt{1537}}{126} when ± is plus. Add -5 to \sqrt{1537}.

$$x=\frac{\sqrt{1537}-5}{126}$$

Now solve the equation x=\frac{-5±\sqrt{1537}}{126} when ± is minus. Subtract \sqrt{1537} from -5.

$$x=\frac{-\sqrt{1537}-5}{126}$$

The equation is now solved.

$$x=\frac{\sqrt{1537}-5}{126} x=\frac{-\sqrt{1537}-5}{126}$$

Substitute \frac{\sqrt{1537}-5}{126} for x in the equation 8x=\sqrt{x^{2}-5x+6}.

$$8\times \frac{\sqrt{1537}-5}{126}=\sqrt{\left(\frac{\sqrt{1537}-5}{126}\right)^{2}-5\times \frac{\sqrt{1537}-5}{126}+6}$$

Simplify. The value x=\frac{\sqrt{1537}-5}{126} satisfies the equation.

$$\frac{4}{63}\times 1537^{\frac{1}{2}}-\frac{20}{63}=\frac{4}{63}\times 1537^{\frac{1}{2}}-\frac{20}{63}$$

Substitute \frac{-\sqrt{1537}-5}{126} for x in the equation 8x=\sqrt{x^{2}-5x+6}.

$$8\times \frac{-\sqrt{1537}-5}{126}=\sqrt{\left(\frac{-\sqrt{1537}-5}{126}\right)^{2}-5\times \frac{-\sqrt{1537}-5}{126}+6}$$

Simplify. The value x=\frac{-\sqrt{1537}-5}{126} does not satisfy the equation because the left and the right hand side have opposite signs.

$$-\frac{4}{63}\times 1537^{\frac{1}{2}}-\frac{20}{63}=\frac{4}{63}\times 1537^{\frac{1}{2}}+\frac{20}{63}$$

Equation 8x=\sqrt{x^{2}-5x+6} has a unique solution.

$$x=\frac{\sqrt{1537}-5}{126}$$