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Question

Question: How do you solve the equation \({{x}^{2}}-3x-7=0\)...

How do you solve the equation x23x7=0{{x}^{2}}-3x-7=0

Explanation

Solution

To solve this equation we will try to create a complete square in the given equation. To do so we will first make the coefficient of x2{{x}^{2}} as 1 if it is not 1. Then we will add and subtract the term (b2a)2{{\left( \dfrac{b}{2a} \right)}^{2}} on both sides and hence use the formula (a+b)2=a2+2ab+b2{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} to simplify the equation then we will take square root of the obtained equation and solve the linear equation to find x.

Complete step-by-step solution:
Now the given equation is a quadratic equation of the form ax2+bx+c=0a{{x}^{2}}+bx+c=0
Comparing the equation with the general equation we get a = 1, b = - 3 and c = - 7.
To solve this equation we will use the method of completing squares.
Now to use this method we need the coefficient of x2{{x}^{2}} to be 1.
Since in the given equation we already have the coefficient as 1 we will proceed with the method.
Now first we will add and subtract the equation with the term (b2a)2{{\left( \dfrac{b}{2a} \right)}^{2}}
Hence adding and subtracting (32)2=94{{\left( \dfrac{-3}{2} \right)}^{2}}=\dfrac{9}{4} to the equation we get,
x23x+(94)(94)7=0\Rightarrow {{x}^{2}}-3x+\left( \dfrac{9}{4} \right)-\left( \dfrac{9}{4} \right)-7=0
Now we know that (a+b)2=a2+2ab+b2{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} Hence using this we get the equation as,
(x32)2=94+7 (x32)2=9+7×44 (x32)2=374 \begin{aligned} & \Rightarrow {{\left( x-\dfrac{3}{2} \right)}^{2}}=\dfrac{9}{4}+7 \\\ & \Rightarrow {{\left( x-\dfrac{3}{2} \right)}^{2}}=\dfrac{9+7\times 4}{4} \\\ & \Rightarrow {{\left( x-\dfrac{3}{2} \right)}^{2}}=\dfrac{37}{4} \\\ \end{aligned}
Now taking square root on both the sides we get,

& \Rightarrow \left( x-\dfrac{3}{2} \right)=\left( \pm \sqrt{\dfrac{37}{4}} \right) \\\ & \Rightarrow \left( x-\dfrac{3}{2} \right)=\pm \dfrac{\sqrt{37}}{2} \\\ \end{aligned}$$ $$\begin{aligned} & \Rightarrow x=\dfrac{3}{2}\pm \dfrac{\sqrt{37}}{2} \\\ & \Rightarrow x=\dfrac{3\pm \sqrt{37}}{2} \\\ \end{aligned}$$ **Hence the solution of the given equation is $x=\dfrac{3\pm \sqrt{37}}{2}$.** **Note:** Now note that the while solving the equation by complete square method the coefficient of ${{x}^{2}}$ must be 1. If it is not equal to one then we divide the whole equation by a to get the coefficient as 1. Now to solve the equation we can also use the formula to find roots of quadratic equation. The roots of the equation of the form $a{{x}^{2}}+bx+c=0$ is given by $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .