Question

Solve 5x 2 - 7x + 2 = 0 by completing the square. What are the solutions?

Answer 1

Answer:

$x=\frac{2}{5}$ or $x=1$

Step-by-step explanation:

The given quadratic equation is

$5x^2-7x+2=0$

Group the constant terms on the right hand side.

$5x^2-7x=0-2$

$5x^2-7x=-2$

Divide through by 5.

$x^2-\frac{7}{5}x=-\frac{2}{5}$

Add the square of half the coefficient of x., which is $(\frac{1}{2}\times- \frac{7}{5})^2=\frac{49}{100}$ to both sides of the equation.

$x^2-\frac{7}{5}x+\frac{49}{100}=-\frac{2}{5}+\frac{49}{100}$

The left hand side is now a perfect square.

$(x-\frac{7}{10})^2=\frac{9}{100}$

Take the square root of both sides;

$(x-\frac{7}{10})=\pm \sqrt{\frac{9}{100}}$

$x-\frac{7}{10}=\pm \frac{3}{10}$

$x=\frac{7}{10}\pm \frac{3}{10}$

$x=\frac{7-3}{10}$ or $x=\frac{7+3}{10}$

$x=\frac{4}{10}$ or $x=\frac{10}{10}$

$x=1$ or $x=\frac{2}{5}$

Answer 2

Answer:

$x_1=1\\\\x_2=\frac{2}{5}$

Step-by-step explanation:

- You must divide the equation by 5:

$x^{2}-\frac{7}{5}x+\frac{2}{5}=0$

- Add and subtract $(\frac{\frac{7}{5}}{2})^2$:

 $x^{2}-\frac{7}{5}x+(\frac{7}{10})^2+\frac{2}{5}-(\frac{7}{10})^2=0$

Therefore, you obtain:

$(x-\frac{7}{10})^2-0.09=0$

-add 0.09} to both sides:

$(x-\frac{7}{10})^2=0.09$

- Apply square root to both sides and solve for x:

$\sqrt{(x-\frac{7}{10})^2}=\sqrt{\0.09}\\x-\frac{7}{10}=\sqrt{0.09}\\\\x_1=\frac{7}{10}+\sqrt{0.09}=1\\\\x_2=-\frac{7}{10}-\sqrt{0.09}=\frac{2}{5}$