Solve 1+a^2+a^2-25a+34=00

$$1 + a ^ { 2 } + a ^ { 2 } - 25 a + 34 = 00$$

Solve for a

$$a = \frac{\sqrt{345} + 25}{4} \approx 10.893543905$$

$$a = \frac{25 - \sqrt{345}}{4} \approx 1.606456095$$

Steps Using the Quadratic Formula

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

$$1+2a^{2}-25a+34=0\times 0$$

Add 1 and 34 to get 35.

$$35+2a^{2}-25a=0\times 0$$

Multiply 0 and 0 to get 0.

$$35+2a^{2}-25a=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$2a^{2}-25a+35=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -25 for b, and 35 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-\left(-25\right)±\sqrt{\left(-25\right)^{2}-4\times 2\times 35}}{2\times 2}$$

Square -25.

$$a=\frac{-\left(-25\right)±\sqrt{625-4\times 2\times 35}}{2\times 2}$$

Multiply -4 times 2.

$$a=\frac{-\left(-25\right)±\sqrt{625-8\times 35}}{2\times 2}$$

Multiply -8 times 35.

$$a=\frac{-\left(-25\right)±\sqrt{625-280}}{2\times 2}$$

Add 625 to -280.

$$a=\frac{-\left(-25\right)±\sqrt{345}}{2\times 2}$$

The opposite of -25 is 25.

$$a=\frac{25±\sqrt{345}}{2\times 2}$$

Multiply 2 times 2.

$$a=\frac{25±\sqrt{345}}{4}$$

Now solve the equation a=\frac{25±\sqrt{345}}{4} when ± is plus. Add 25 to \sqrt{345}.

$$a=\frac{\sqrt{345}+25}{4}$$

Now solve the equation a=\frac{25±\sqrt{345}}{4} when ± is minus. Subtract \sqrt{345} from 25.

$$a=\frac{25-\sqrt{345}}{4}$$

The equation is now solved.

$$a=\frac{\sqrt{345}+25}{4} a=\frac{25-\sqrt{345}}{4}$$

Steps for Completing the Square

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

$$1+2a^{2}-25a+34=0\times 0$$

Add 1 and 34 to get 35.

$$35+2a^{2}-25a=0\times 0$$

Multiply 0 and 0 to get 0.

$$35+2a^{2}-25a=0$$

Subtract 35 from both sides. Anything subtracted from zero gives its negation.

$$2a^{2}-25a=-35$$

Divide both sides by 2.

$$\frac{2a^{2}-25a}{2}=-\frac{35}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$a^{2}-\frac{25}{2}a=-\frac{35}{2}$$

Divide -\frac{25}{2}, the coefficient of the x term, by 2 to get -\frac{25}{4}. Then add the square of -\frac{25}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}-\frac{25}{2}a+\left(-\frac{25}{4}\right)^{2}=-\frac{35}{2}+\left(-\frac{25}{4}\right)^{2}$$

Square -\frac{25}{4} by squaring both the numerator and the denominator of the fraction.

$$a^{2}-\frac{25}{2}a+\frac{625}{16}=-\frac{35}{2}+\frac{625}{16}$$

Add -\frac{35}{2} to \frac{625}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$a^{2}-\frac{25}{2}a+\frac{625}{16}=\frac{345}{16}$$

Factor a^{2}-\frac{25}{2}a+\frac{625}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(a-\frac{25}{4}\right)^{2}=\frac{345}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-\frac{25}{4}\right)^{2}}=\sqrt{\frac{345}{16}}$$

Simplify.

$$a-\frac{25}{4}=\frac{\sqrt{345}}{4} a-\frac{25}{4}=-\frac{\sqrt{345}}{4}$$

Add \frac{25}{4} to both sides of the equation.

$$a=\frac{\sqrt{345}+25}{4} a=\frac{25-\sqrt{345}}{4}$$