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Question: In the given question, we have been asked to find the value of ‘x’ by solving the given equation i.e...

In the given question, we have been asked to find the value of ‘x’ by solving the given equation i.e. 2x2−6x−20=02{{x}^{2}}-6x-20=0 using the completing the square method. Completing the square method is used to solve the quadratic equation by converting the form of the equation so that it will become a perfect trinomial.

Explanation

Solution

In the given question, we have been asked to find the value of ‘x’ by solving the given equation i.e. 2x2−6x−20=02{{x}^{2}}-6x-20=0using the completing the square method. Completing the square method is used to solve the quadratic equation by converting the form of the equation so that it will become a perfect trinomial.

Formula used:
a2−2ab+b2=(a−b)2{{a}^{2}}-2ab+{{b}^{2}}={{\left( a-b \right)}^{2}}

Complete step by step solution:
We have given that,
⇒2x2−6x−20=0\Rightarrow 2{{x}^{2}}-6x-20=0
Dividing both the side of the equation by 2, we get
⇒x2−3x−10=0\Rightarrow {{x}^{2}}-3x-10=0
Adding 10 to both the sides of the equation, we get
⇒x2−3x−10+10=0+10\Rightarrow {{x}^{2}}-3x-10+10=0+10
Simplifying the above equation, we get
⇒x2−3x=10\Rightarrow {{x}^{2}}-3x=10
Now, we completing the square, we add (32)2{{\left( \dfrac{3}{2} \right)}^{2}}to both the sides of the equation, we get
⇒x2−3x+(32)2=10+(32)2\Rightarrow {{x}^{2}}-3x+{{\left( \dfrac{3}{2} \right)}^{2}}=10+{{\left( \dfrac{3}{2} \right)}^{2}}
Simplifying the above equation, we get
⇒x2−3x+(32)2=101+(94)\Rightarrow {{x}^{2}}-3x+{{\left( \dfrac{3}{2} \right)}^{2}}=\dfrac{10}{1}+\left( \dfrac{9}{4} \right)
Taking the LCM of 4 and 1 on the right side of the equation, we get
LCM of 4 and 1 is equal to 4.
⇒x2−3x+(32)2=494\Rightarrow {{x}^{2}}-3x+{{\left( \dfrac{3}{2} \right)}^{2}}=\dfrac{49}{4}
As we know that, a2−2ab+b2=(a−b)2{{a}^{2}}-2ab+{{b}^{2}}={{\left( a-b \right)}^{2}}
Therefore,
⇒(x−32)2=494\Rightarrow {{\left( x-\dfrac{3}{2} \right)}^{2}}=\dfrac{49}{4}
Transposing the power 2 on the right side of the equation, we get
⇒x−32=494\Rightarrow x-\dfrac{3}{2}=\sqrt{\dfrac{49}{4}}
As we know that 49=±7\sqrt{49}=\pm 7 and 4=±2\sqrt{4}=\pm 2
Simplifying the above, we get
⇒x−32=±72\Rightarrow x-\dfrac{3}{2}=\pm \dfrac{7}{2}
Adding 32\dfrac{3}{2}to both the sides of the equation, we get
⇒x−32+32=±72+32\Rightarrow x-\dfrac{3}{2}+\dfrac{3}{2}=\pm \dfrac{7}{2}+\dfrac{3}{2}
Simplifying the numbers, we get
⇒x=±72+32\Rightarrow x=\pm \dfrac{7}{2}+\dfrac{3}{2}
Now, we have
⇒x=72+32=102=5\Rightarrow x=\dfrac{7}{2}+\dfrac{3}{2}=\dfrac{10}{2}=5 or x=−72+32=−42=−2x=-\dfrac{7}{2}+\dfrac{3}{2}=- \dfrac{4}{2}=-2
⇒x=5 or −2\Rightarrow x=5\ or\ -2
Therefore, the possible values of ‘x’ are 5 and -2.

Note: While solving these types of questions, students should carefully observe when considering the terms, which one is the ‘a’ and the ‘b’. To check whether the obtained possible values are correct or not, we can verify the result by solving the quadratic equation with the roots of the quadratic equation formula. Standard form of quadratic equation; ax2+bx+c=0a{{x}^{2}}+bx+c=0, then the roots of the quadratic equation is given by, x=−b±b2−4ac2ax=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}.