Question

Some examples of related to complete square method

Answer 1

Step-by-step explanation:

1) x² – 4x – 8

2) x² + 6x + 2 = 0

Answer 2

Answer:

1. $\sf\:x^{2}\:+\:2x\:-\:5\:=\:0$

2. $\sf\:5x^{2}\:-\:4x\:-\:7\:=\:0$

Step-by-step-explanation:

1. $\sf\:x^{2}\:+\:2x\:-\:5\:=\:0$

$\sf\:x^{2}\:+\:2x\:-\:5\:=\:0$

$\sf\:Comparing\:x^{2}\:+\:2x\:with\:x^{2}\:+\:2xy,\\\\\sf\:2xy\:=\:2x\\\\\therefore\sf\:y\:=\:1\\\\\therefore\sf\:y^{2}\:=\:1\\\\\therefore\sf\:x^{2}\:+\:2x\:+\:1\:is\:a\:perfect\:square\:trinomial.\\\\\therefore\sf\:x^{2}\:+\:2x\:-\:5\:=\:0\\\\\therefore\sf\:x^{2}\:+\:2x\:+\:1\:-\:1\:-\:5\:=\:0\\\\\implies\sf\:(\:x\:+\:1\:)^{2}\:-\:6\:=\:0\\\\\implies\sf\:(\:x\:+\:1\:)^{2}\:-\:(\:\sqrt{6}\:)^{2}\:=\:0\\\\\implies\sf\:(\:x\:+\:1\:+\:\sqrt{6}\:)\:(\:x\:+\:1\:-\:\sqrt{6}\:)\:=\:0\:\:\:-\:-\:-\:[\:a^{2}\:-\:b^{2}\:=\:(\:a\:+\:b\:)\:(\:a\:-\:b\:)\:]\\\\\implies\sf\:x\:+\:1\:+\:\sqrt{6}\:\:\:or\:\:\:x\:+\:1\:-\:\sqrt{6}\:=\:0\\\\\implies\boxed{\red{\sf\:x\:=\:-\:1\:-\:\sqrt{6}}}\:\:\:\sf\:or\:\:\:\boxed{\red{\sf\:x\:=\:-\:1\:+\:\sqrt{6}}}$

2. $\sf\:5x^{2}\:-\:4x\:-\:7\:=\:0$.

$\sf\:5x^{2}\:-\:4x\:-\:7\:=\:0\\\\\therefore\sf\:x^{2}\:-\:\frac{4x}{5}\:-\:\frac{7}{5}\:=\:0\:\:\:-\:-\:-\:[\:Dividing\:both\:sides\:by\:5\:]\\\\\sf\:Comparing\:x^{2}\:-\:\frac{4x}{5}\:with\:x^{2}\:-\:2xy\\\\\therefore\sf\:-\:2xy\:=\:-\:\frac{4x}{5}\\\\\therefore\sf\:y\:=\:\frac{2}{5}\\\\\therefore\sf\:y^{2}\:=\:\frac{4}{25}\\\\\therefore\sf\:x^{2}\:-\:\frac{4x}{5}\:+\:\frac{4}{25}\:is\:a\:perfect\:square\:trinomial.\\\\\therefore\sf\:x^{2}\:-\:\frac{4x}{5}\:-\:\frac{7}{5}\:=\:0\\\\\implies\sf\:x^{2}\:-\:\frac{4x}{5}\:+\:\frac{4}{25}\:-\:\frac{4}{25}\:-\:\frac{7}{5}\:=\:0\\\\\implies\sf\:(\:x\:-\:\frac{2}{5}\:)^{2}\:-\:(\:\frac{4}{25}\:+\:\frac{7}{5}\:)\:=\:0\\\\\implies\sf\:(\:x\:-\:\frac{2}{5}\:)^{2}\:-\:(\:\dfrac{\:4\:+\:35\:}{25}\:)\:=\:0\\\\\implies\sf\:(\:x\:-\:\frac{2}{5}\:)^{2}\:-\:(\:\frac{39}{25}\:)\:=\:0\\\\\implies\sf\:(\:x\:-\:\frac{2}{5}\:)^{2}\:-\:(\:\frac{\sqrt{39}}{5}\:)^{2}\:=\:0\\\\\implies\sf\:(\:x\:-\:\frac{2}{5}\:+\:\frac{\sqrt{39}}{5}\:)\:(\:x\:-\:\frac{2}{5}\:-\:\frac{\sqrt{39}}{5}\:)\:=\:0\:\:\:-\:-\:-\:[\:a^{2}\:-\:b^{2}\:=\:(\:a\:+\:b\:)\:(\:a\:-\:b\:)\:]\\\\\implies\sf\:x\:-\:\frac{2}{5}\:+\:\frac{\sqrt{39}}{5}\:\:\:or\:\:\:x\:-\:\frac{2}{5}\:-\:\frac{\sqrt{39}}{5}\:=\:0\\\\\implies\sf\:x\:=\:\frac{2}{5}\:-\:\frac{\sqrt{39}}{5}\:\:\:or\:\:\:x\:=\:\frac{2}{5}\:+\:\frac{\sqrt{39}}{5}\\\\\implies\boxed{\red{\sf\:x\:=\:\dfrac{2\:-\:\sqrt{39}}{5}}}\:\:\:\sf\:or\:\:\:\boxed{\red{\sf\:x\:=\:\dfrac{2\:+\:\sqrt{39}}{5}}}$

Additional Information:

1. Quadratic Equation :

An equation having a degree '2' is called quadratic equation.

The general form of quadratic equation is

ax² + bx + c = 0

Where, a, b, c are real numbers and a ≠ 0.

2. Roots of Quadratic Equation:

The roots means nothing but the value of the variable given in the equation.

3. Methods of solving quadratic equation:

There are mainly three methods to solve or find the roots of the quadratic equation.

A) Factorization method

B) Completing square method

C) Formula method

4. Solution of Quadratic Equation by Completing Square Method:

1. Write the given quadratic equation in the form $\sf\:ax^{2}\:+\:bx\:+\:c\:=\:0$.

2. Consider the first two terms on LHS and find the third suitable square term.

3. It will make a perfect square polynomial.

4. Add and subtract the square term to given equation.

5. Write the square of the first three terms and last two terms.

6. Factorize that term and equate to zero.

7. Find the value of variable.