Question

Solve \[{{x}^{2}}+7x=7\] and give your answer correct to two decimal places.

Answer 1

Hint: We have the equation, \[{{x}^{2}}+7x=7\] . Shift 7 from RHS of the equation to LHS of the given equation. Then, we can see that the given equation is a quadratic equation. We know the formula to find the roots, \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] . Here a=1, b=7, and c=-7. Now, using the formula we can get both roots of the given equation. We know that, \[\sqrt{77}=8.77496\].

Complete step-by-step answer:
According to the question we have,
\[{{x}^{2}}+7x=7\] ……………..(1)
Now, in equation (1), move the constant term 7 to the LHS of the equation.
\[{{x}^{2}}+7x=7\]
\[\Rightarrow {{x}^{2}}+7x-7=0\] ………………………..(2)
Now, we can see that equation (2) is a quadratic equation.
We know that if a quadratic equation, \[a{{x}^{2}}+bx+c=0\] then its root is given by
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] …………………………(3)
Comparing, \[a{{x}^{2}}+bx+c=0\] and \[{{x}^{2}}+7x-7=0\] , we get
a=1, b=7, and c=-7.
Now, putting the values of a, b, and c in equation (3), we get
\[x=\dfrac{-7\pm \sqrt{{{7}^{2}}-4.1.(-7)}}{2.1}\]
\[\begin{align}
  & \Rightarrow x=\dfrac{-7\pm \sqrt{49+28}}{2} \\
 & \Rightarrow x=\dfrac{-7\pm \sqrt{77}}{2} \\
\end{align}\]
We know that, \[\sqrt{77}=8.77496\] . Now putting it in the above expression, we get
\[\begin{align}
  & \Rightarrow x=\dfrac{-7\pm 8.77496}{2} \\
 & x=\dfrac{-7-8.77496}{2}=-7.88,x=\dfrac{-7+8.77496}{2}=0.88 \\
\end{align}\]
Hence, the value of x can be -7.88 or 0.88.

Note: We can also solve this question using the perfect square method.
According to the question we have to solve, \[{{x}^{2}}+7x=7\] ……………..(1)
Now, in equation (1), move the constant term 7 to the LHS of the equation.
\[{{x}^{2}}+7x=7\]
\[\Rightarrow {{x}^{2}}+7x-7=0\] ………………………..(2)
To make a perfect square add and subtract the constant term \[\dfrac{49}{4}\] in LHS of the equation (2).
Now,

\[{{(x)}^{2}}+2.\dfrac{7}{2}x+\dfrac{49}{4}-\dfrac{49}{4}-7=0\]
\[{{(x)}^{2}}+2.\dfrac{7}{2}x+{{\left( \dfrac{7}{2} \right)}^{2}}-\dfrac{49}{4}-7=0\] …………………………..(3)
We know the formula, \[{{a}^{2}}+2.a.b+{{b}^{2}}={{(a+b)}^{2}}\] …………….(4)
Using equation (4), we can write equation (3) as,
\[\begin{align}
  & \Rightarrow {{\left( x+\dfrac{7}{2} \right)}^{2}}=\dfrac{77}{4} \\
 & \Rightarrow \left( x+\dfrac{7}{2} \right)=\sqrt{\dfrac{77}{4}} \\
 & \Rightarrow x=-\dfrac{7}{2}\pm \dfrac{\sqrt{77}}{2} \\
\end{align}\]
We know that, \[\sqrt{77}=8.77496\] . Now putting it in the above expression, we get
\[\begin{align}
  & \Rightarrow x=\dfrac{-7\pm 8.77496}{2} \\
 & x=\dfrac{-7-8.77496}{2}=-7.88,x=\dfrac{-7+8.77496}{2}=0.88 \\
\end{align}\]
Hence, the value of x can be -7.88 or 0.88.