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}

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

x = \frac{6}{4}

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.

Did you find an error?

6. Solving a quadratic equation by completing the square method