Question

How do you solve $2{{x}^{2}}+6x-3=0$ by completing the square ?

Answer 1

Hint: We can convert any quadratic equation $a{{x}^{2}}+bx+c$ to complete square , if we convert $a{{x}^{2}}+bx+c$ into complete square and solve for x we will ultimately get the root for quadratic formula that is $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .

Complete step by step answer:
The equation given in the question is $2{{x}^{2}}+6x-3=0$
We can see it is a quadratic equation, let’s solve it by completing square
The coefficient of ${{x}^{2}}$ in the equation $2{{x}^{2}}+6x-3=0$ is 2, first divide the whole equation by 2
${{x}^{2}}+3x-\dfrac{3}{2}$ = 0
We can write the equation as ${{x}^{2}}+2\times \dfrac{3}{2}x-\dfrac{3}{2}$
We know that ${{\left( a-b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$
So if there is a term ${{\left( \dfrac{3}{2} \right)}^{2}}$ in the equation we can write it as a square .
We add ${{\left( \dfrac{3}{2} \right)}^{2}}$ in LHS and RHS
${{x}^{2}}+2\times \dfrac{3}{2}x-\dfrac{3}{2}+\dfrac{9}{4}=\dfrac{9}{4}$
Now we can write $\Rightarrow {{\left( x+\dfrac{3}{2} \right)}^{2}}-\dfrac{3}{2}=\dfrac{9}{4}$
We can add $\dfrac{3}{2}$ to both LHS and RHS
$\Rightarrow {{\left( x+\dfrac{3}{2} \right)}^{2}}=\dfrac{9}{4}+\dfrac{3}{2}$
$\Rightarrow {{\left( x+\dfrac{3}{2} \right)}^{2}}=\dfrac{15}{4}$
$\Rightarrow \left( x+\dfrac{3}{2} \right)=\dfrac{\pm \sqrt{15}}{2}$
Now subtracting $\dfrac{3}{2}$ to LHS and RHS
$\Rightarrow x=\dfrac{-3\pm \sqrt{15}}{2}$
So the roots of the equation are $\dfrac{-3+\sqrt{15}}{2}$ and $\dfrac{-3-\sqrt{15}}{2}$

Note:
The derivation of the formula for roots of a quadratic equation is done by method of completing square , the result of completing square of $a{{x}^{2}}+bx+c$ = 0 is ${{\left( x+\dfrac{b}{2a} \right)}^{2}}=\dfrac{{{b}^{2}}-4ac}{4{{a}^{2}}}$
If we observe the denominator of $\dfrac{{{b}^{2}}-4ac}{4{{a}^{2}}}$ is always positive because it is square of 2a. LHS of ${{\left( x+\dfrac{b}{2a} \right)}^{2}}=\dfrac{{{b}^{2}}-4ac}{4{{a}^{2}}}$ is square number, so if ${{b}^{2}}-4ac$is less than 0 then $\dfrac{{{b}^{2}}-4ac}{4{{a}^{2}}}$ will be negative so the roots will be complex and if ${{b}^{2}}-4ac$ greater than 0, then roots are real. If a is positive then minimum value of $a{{x}^{2}}+bx+c$ is $\dfrac{4ac-{{b}^{2}}}{4{{a}^{2}}}$ and if a is negative then maximum value of $a{{x}^{2}}+bx+c$ is $\dfrac{4ac-{{b}^{2}}}{4{{a}^{2}}}$ .