Solving a quadratic equation by completing the square method — lesson. Mathematics State Board, Class 10.

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Let us see the procedure to solve the quadratic equation by completing the square.

Step 1: Write the given equation in standard form

$ax^2 + bx + c = 0$ .

Step 2: Make sure the coefficient of

$x^2$ is

$a = 1$ . If not, make it by dividing the equation by

$a$ .

Step 3: Move the constant term to the right-hand side of the equation.

Step 4: Add the square of one-half of the coefficient of

$x$ to both sides. [That is, add

$\left(\frac{b}{2}\right)^2$ .]

Step 5: Make the equation a complete square and simplify the right-hand side.

Step 6: Solve for

$x$ by taking the square root on both sides.

Example:

Find the root of

$2x^2 + 7x - 15 = 0$ by the method of completing the square.

Solution:

The given equation is

$2x^2 + 7x - 15 = 0$ .

Here, the coefficient of

$x^2$ is

$2$ . So, divide the equation by

$2$ .

$\frac{2}{2} x^{2} + \frac{7}{2} x - \frac{15}{2} = \frac{0}{2}$

$x^{2} + \frac{7}{2} x - \frac{15}{2} = 0$

Move the constant term to the right hand side of the equation.

$x^{2} + \frac{7}{2} x = \frac{15}{2}$

Add the square of one half of the coefficient of

$x$ to both sides.

$x^{2} + \frac{7}{2} x + {(\frac{7}{4})}^{2} = \frac{15}{2} + {(\frac{7}{4})}^{2}$

$x^{2} + 2 \times \frac{7}{4} x + {(\frac{7}{4})}^{2} = \frac{15}{2} + \frac{49}{16}$

Left hand side equation reminds the identity

$(a + b)^2 = a^2 + 2ab + b^2$ .

${(x + \frac{7}{4})}^{2} = \frac{120 + 49}{16}$

${(x + \frac{7}{4})}^{2} = \frac{169}{16}$

Taking square root on both sides.

$x + \frac{7}{4} = \pm \frac{13}{4}$

$x = \frac{13}{4} - \frac{7}{4}$ or

$x = - \frac{13}{4} - \frac{7}{4}$

$x = \frac{6}{4}$ or

$x = \frac{- 20}{4}$

$x =$

$\frac{3}{2}$ or

$x = -5$

Therefore, the roots of the given equation are

$\frac{3}{2}$ and

$-5$ .

2. Find the roots of the equation

$x^2 + x + 2 = 0$ by completing the square method.

Solution:

The given equation is

$x^2 + x + 2 = 0$

Here, the coefficient of

$x^2$ is

$1$ .

$x^2 + x + 2 = 0$

Move the constant term to the right hand side of the equation.

$x^2 + x = -2$

Add the square of one half of the coefficient of

$x$ to both sides.

$x^{2} + x + {(\frac{1}{2})}^{2} = - 2 + {(\frac{1}{2})}^{2}$

$x^{2} + 2 \times \frac{1}{2} x + {(\frac{1}{2})}^{2} = - 2 + \frac{1}{4}$

${(x + \frac{1}{2})}^{2} = - \frac{7}{4}$

$< 0$

The above equation cannot be possible because the square of any number cannot be negative.

Therefore, the given equation has no real roots.

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6. Solving a quadratic equation by completing the square method

Theory: