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Question

Question: How do you solve \(2{{x}^{2}}+6x=10\) ?...

How do you solve 2x2+6x=102{{x}^{2}}+6x=10 ?

Explanation

Solution

We can solve for the solution of the given equation by applying the complete square method. We will add numbers on both the sides after rearranging some terms to have a complete square in the form of (x+a)2{{\left( x+a \right)}^{2}} . Now, further simplifying the equation we will be able to find the solution of the equation.

Complete step-by-step solution:
The given equation is
2x2+6x=102{{x}^{2}}+6x=10
We now divide both sides of the above equation by 22 as
x2+3x=5\Rightarrow {{x}^{2}}+3x=5
We rewrite the equation as
x2+232x=5....expression1\Rightarrow {{x}^{2}}+2\cdot \dfrac{3}{2}x=5....\text{expression}1
To have a complete square in the left hand side, let’s take the square (x+a)2{{\left( x+a \right)}^{2}} for comparison.
We know that, (x+a)2=x2+2ax+a2....expression2{{\left( x+a \right)}^{2}}={{x}^{2}}+2\cdot a\cdot x+{{a}^{2}}....\text{expression2}
As, the first two terms in the expression (x2+2ax+a2)\left( {{x}^{2}}+2\cdot a\cdot x+{{a}^{2}} \right) are x2{{x}^{2}} and 2x2x , we compare the left-hand side of expression1\text{expression}1 with the right-hand side of expression2\text{expression2} and get the value of aa as a=32a=\dfrac{3}{2} .
Hence, to get the square term (x+32)2{{\left( x+\dfrac{3}{2} \right)}^{2}} we add 94\dfrac{9}{4} to the both sides of expression1\text{expression}1
x2+232x+94=5+94\Rightarrow {{x}^{2}}+2\cdot \dfrac{3}{2}x+\dfrac{9}{4}=5+\dfrac{9}{4}
x2+232x+94=294\Rightarrow {{x}^{2}}+2\cdot \dfrac{3}{2}x+\dfrac{9}{4}=\dfrac{29}{4}
The above equation can be also written as
(x+32)2=294\Rightarrow {{\left( x+\dfrac{3}{2} \right)}^{2}}=\dfrac{29}{4}
Now, taking square root on both the sides of the equation, we get
x+32=±292\Rightarrow x+\dfrac{3}{2}=\pm \dfrac{\sqrt{29}}{2}
Further simplifying the above equation, we get
x=29232\Rightarrow x=\dfrac{\sqrt{29}}{2}-\dfrac{3}{2}
x=2932\Rightarrow x=\dfrac{\sqrt{29}-3}{2}
And x=29232\Rightarrow x=-\dfrac{\sqrt{29}}{2}-\dfrac{3}{2}
x=29+32\Rightarrow x=-\dfrac{\sqrt{29}+3}{2}
Therefore, the solution of the equation 2x2+6x=102{{x}^{2}}+6x=10 are x=29+32x=-\dfrac{\sqrt{29}+3}{2} and x=2932x=\dfrac{\sqrt{29}-3}{2}.

Note: We have to keep in mind while simplifying the last part of the solution we have to take both the positive and negative signs of 292\dfrac{\sqrt{29}}{2} , otherwise we will not be getting both the solutions. The given problem can also be solved by other methods, such as by factoring the left-hand side of the equation and equating both the factors individually to zero. Also, it can be done by using directly Sridhar Acharya formula to find the solution, which is x=b±b24ac2ax=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} .