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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.

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

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

Left hand side equation reminds the identity \((a + b)^2 = a^2 + 2ab + b^2\).

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

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

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.

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

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

${\left(x+\frac{1}{2}\right)}^{2}=-\frac{7}{4}$ \(< 0\)

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

**Therefore, the given equation has no real roots**.