Question

How do you solve this by completing the square for $2{{x}^{2}}+7x-15=0$?

Answer 1

To solve the given equation by using completing the square method first we will transform the equation such that the constant term remains alone at the right side. Then we add or subtract a number from the given equation to make the constant term a perfect square. Then factor the left terms and solve for x.

Complete step-by-step solution:
We have been given an equation $2{{x}^{2}}+7x-15=0$.
We have to solve the given equation by completing the square method.
Now, first let us add 15 to the given equation we will get
$\begin{aligned} & \Rightarrow 2{{x}^{2}}+7x-15=0+15 \\\ & \Rightarrow 2{{x}^{2}}+7x=15 \\\ \end{aligned}$
We need to make the coefficient of ${{x}^{2}}$ as 1 we will divide the whole equation by 2. Then we will get
$\begin{aligned} & \Rightarrow \dfrac{2{{x}^{2}}}{2}+\dfrac{7x}{2}=\dfrac{15}{2} \\\ & \Rightarrow {{x}^{2}}+\dfrac{7x}{2}=\dfrac{15}{2} \\\ \end{aligned}$
Now, we need to add ${{\left( \dfrac{b}{2a} \right)}^{2}}$ to both sides of the equation to make the constant term a perfect square we get

& \Rightarrow {{\left( \dfrac{b}{2a} \right)}^{2}}={{\left( \dfrac{\dfrac{7}{2}}{2\times 1} \right)}^{2}} \\\ & \Rightarrow {{\left( \dfrac{\dfrac{7}{2}}{2} \right)}^{2}} \\\ & \Rightarrow {{\left( \dfrac{7}{2}\times \dfrac{1}{2} \right)}^{2}} \\\ & \Rightarrow \dfrac{49}{16} \\\ \end{aligned}$$ Now, add $\dfrac{49}{16}$ to both sides of the equation we get $\Rightarrow {{x}^{2}}+\dfrac{7}{2}x+\dfrac{49}{16}=\dfrac{15}{2}+\dfrac{49}{16}$ Now, simplifying the above equation we will get $\begin{aligned} & \Rightarrow {{x}^{2}}+\dfrac{7}{2}x+\dfrac{49}{16}=\dfrac{120+49}{16} \\\ & \Rightarrow {{x}^{2}}+\dfrac{7}{2}x+\dfrac{49}{16}=\dfrac{169}{16} \\\ \end{aligned}$ Now, we can simplify the LHS to make it the perfect square we will get $\begin{aligned} & \Rightarrow {{x}^{2}}+2\times \left( \dfrac{7}{4} \right)x+{{\left( \dfrac{7}{4} \right)}^{2}}=\dfrac{169}{16} \\\ & \Rightarrow {{\left( x+\dfrac{7}{4} \right)}^{2}}={{\left( \dfrac{13}{4} \right)}^{2}} \\\ \end{aligned}$ Now, taking the square root both sides and solving further we will get $\begin{aligned} & \Rightarrow x+\dfrac{7}{4}=\pm \dfrac{13}{4} \\\ & \Rightarrow x=\pm \dfrac{13}{4}-\dfrac{7}{4} \\\ \end{aligned}$ So we get two values of x as $\begin{aligned} & \Rightarrow x=\dfrac{-13-7}{4},x=\dfrac{13-7}{4} \\\ & \Rightarrow x=\dfrac{-20}{4},x=\dfrac{6}{4} \\\ & \Rightarrow x=-5,x=\dfrac{3}{2} \\\ \end{aligned}$ **Hence the solutions of the given equation are $x=-5,\dfrac{3}{2}$ .** **Note:** If method is not specified in the question then we can solve the quadratic equation by any one of the methods from completing the square, quadratic formula or by factorization method. The possibility of mistake while solving the question is that after taking the square root on both sides students simply solve the equation as $\begin{aligned} & \Rightarrow x+\dfrac{7}{4}=\dfrac{13}{4} \\\ & \Rightarrow x=\dfrac{13}{4}-\dfrac{7}{4} \\\ & \Rightarrow x=\dfrac{13-7}{4} \\\ & \Rightarrow x=\dfrac{6}{4} \\\ & \Rightarrow x=\dfrac{3}{2} \\\ \end{aligned}$ It gives only one solution, which is incomplete. So considering both signs is necessary.