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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.

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

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

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.

$x+\frac{7}{4}=±\frac{13}{4}$

$x=\frac{13}{4}-\frac{7}{4}$ 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.

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

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

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

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

Therefore, the given equation has no real roots.