
How do you solve $2{{x}^{2}}+3x-17=0$ by completing the square?
Answer
548.4k+ views
Hint: In this question, we have to find the value of x. Since the equation is a type quadratic form, thus we get two values of x. Therefore, we will use completing the square method, to get the solution. We start solving this problem, by adding 17 on both sides of the equation and then dividing 2 on both sides of the equation. After the necessary calculations, we will add ${{\left( \dfrac{3}{4} \right)}^{2}}$ on both sides of the equation. Then, we will create the left-hand side of the equation as a perfect square. In the end, we will take the square root on both sides and make the necessary calculations. Then, we will get two equations, thus we solve them separately, which is the required solution for the problem.
Complete step by step solution:
According to the question, we have to find the value of x.
Thus, we will use the completing the square method to get the solution.
The equation given to us is $2{{x}^{2}}+3x-17=0$ ---------- (1)
We start solving this problem by adding 17 on both sides in the equation (1), we get
$\Rightarrow 2{{x}^{2}}+3x-17+17=0+17$
As we know, the same terms with opposite signs cancel out each other, thus we get
$\Rightarrow 2{{x}^{2}}+3x=17$
Now, we will make the coefficient of ${{x}^{2}}$ as 1, thus we will divide 2 on both sides in the above equation, we get
$\Rightarrow \dfrac{2}{2}{{x}^{2}}+\dfrac{3}{2}x=\dfrac{17}{2}$
On further simplification, we get
$\Rightarrow {{x}^{2}}+\dfrac{3}{2}x=\dfrac{17}{2}$
Now, we will add ${{\left( \dfrac{b}{2} \right)}^{2}}$on both sides of the equation, here $b=\dfrac{3}{2}$ , thus we will add ${{\left( \dfrac{3}{4} \right)}^{2}}$on both sides in the above equation, we get
$\Rightarrow {{x}^{2}}+\dfrac{3}{2}x+{{\left( \dfrac{3}{4} \right)}^{2}}=\dfrac{17}{2}+{{\left( \dfrac{3}{4} \right)}^{2}}$
On further solving the above equation, we get
$\Rightarrow {{x}^{2}}+\dfrac{3}{2}x+{{\left( \dfrac{3}{4} \right)}^{2}}=\dfrac{17}{2}+\dfrac{9}{16}$
Now, taking the LCM on the right-hand side of the equation, we get
$\Rightarrow {{x}^{2}}+\dfrac{3}{2}x+{{\left( \dfrac{3}{4} \right)}^{2}}=\dfrac{136+9}{16}$
$\Rightarrow {{x}^{2}}+\dfrac{3}{2}x+{{\left( \dfrac{3}{4} \right)}^{2}}=\dfrac{145}{16}$
So, we will make the perfect square on the left-hand side of the above equation, that is we will use the algebraic identity ${{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ , we get
$\Rightarrow {{\left( x+\left( \dfrac{3}{4} \right) \right)}^{2}}=\dfrac{145}{16}$
Now, we will take the square root on both sides in the above equation, we get
$\Rightarrow \sqrt{{{\left( x+\left( \dfrac{3}{4} \right) \right)}^{2}}}=\sqrt{\dfrac{145}{16}}$
On further solving the above equation, we get
$\Rightarrow \left( x+\left( \dfrac{3}{4} \right) \right)=\pm \dfrac{\sqrt{145}}{4}$
Now, we will split (+) and (-) sign to get two new equations separately, that is
$\Rightarrow \left( x+\left( \dfrac{3}{4} \right) \right)=\dfrac{\sqrt{145}}{4}$ -------- (2) and
$\Rightarrow \left( x+\left( \dfrac{3}{4} \right) \right)=-\dfrac{\sqrt{145}}{4}$ ------- (3)
Now, we will solve equation (2), which is
$\Rightarrow \left( x+\left( \dfrac{3}{4} \right) \right)=\dfrac{\sqrt{145}}{4}$
So, we will subtract $\dfrac{3}{4}$ on both sides in the above equation, we get
$\Rightarrow x+\dfrac{3}{4}-\dfrac{3}{4}=\dfrac{\sqrt{145}}{4}-\dfrac{3}{4}$
As we know, the same terms with opposite signs cancel out each other, thus we get
$\Rightarrow x=\dfrac{\sqrt{145}-3}{4}$
Thus, we will solve equation (3), which is
$\Rightarrow \left( x+\left( \dfrac{3}{4} \right) \right)=-\dfrac{\sqrt{145}}{4}$
So, we will subtract $\dfrac{3}{4}$ on both sides in the above equation, we get
$\Rightarrow x+\dfrac{3}{4}-\dfrac{3}{4}=-\dfrac{\sqrt{145}}{4}-\dfrac{3}{4}$
As we know, the same terms with opposite signs cancel out each other, thus we get
$\Rightarrow x=\dfrac{-\sqrt{145}-3}{4}$
Thus, for the equation $2{{x}^{2}}+3x-17=0$, the value of x are $\dfrac{\sqrt{145}-3}{4},\dfrac{-\sqrt{145}-3}{4}$
Note: While solving this problem, do mention all the steps properly to avoid confusion and mathematical errors. Do not forget to split (+) and (-) sign to get an accurate answer. For checking your answer, substitute the value of x on the left-hand side of the equation, then the answer must be equal to 0.
Complete step by step solution:
According to the question, we have to find the value of x.
Thus, we will use the completing the square method to get the solution.
The equation given to us is $2{{x}^{2}}+3x-17=0$ ---------- (1)
We start solving this problem by adding 17 on both sides in the equation (1), we get
$\Rightarrow 2{{x}^{2}}+3x-17+17=0+17$
As we know, the same terms with opposite signs cancel out each other, thus we get
$\Rightarrow 2{{x}^{2}}+3x=17$
Now, we will make the coefficient of ${{x}^{2}}$ as 1, thus we will divide 2 on both sides in the above equation, we get
$\Rightarrow \dfrac{2}{2}{{x}^{2}}+\dfrac{3}{2}x=\dfrac{17}{2}$
On further simplification, we get
$\Rightarrow {{x}^{2}}+\dfrac{3}{2}x=\dfrac{17}{2}$
Now, we will add ${{\left( \dfrac{b}{2} \right)}^{2}}$on both sides of the equation, here $b=\dfrac{3}{2}$ , thus we will add ${{\left( \dfrac{3}{4} \right)}^{2}}$on both sides in the above equation, we get
$\Rightarrow {{x}^{2}}+\dfrac{3}{2}x+{{\left( \dfrac{3}{4} \right)}^{2}}=\dfrac{17}{2}+{{\left( \dfrac{3}{4} \right)}^{2}}$
On further solving the above equation, we get
$\Rightarrow {{x}^{2}}+\dfrac{3}{2}x+{{\left( \dfrac{3}{4} \right)}^{2}}=\dfrac{17}{2}+\dfrac{9}{16}$
Now, taking the LCM on the right-hand side of the equation, we get
$\Rightarrow {{x}^{2}}+\dfrac{3}{2}x+{{\left( \dfrac{3}{4} \right)}^{2}}=\dfrac{136+9}{16}$
$\Rightarrow {{x}^{2}}+\dfrac{3}{2}x+{{\left( \dfrac{3}{4} \right)}^{2}}=\dfrac{145}{16}$
So, we will make the perfect square on the left-hand side of the above equation, that is we will use the algebraic identity ${{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ , we get
$\Rightarrow {{\left( x+\left( \dfrac{3}{4} \right) \right)}^{2}}=\dfrac{145}{16}$
Now, we will take the square root on both sides in the above equation, we get
$\Rightarrow \sqrt{{{\left( x+\left( \dfrac{3}{4} \right) \right)}^{2}}}=\sqrt{\dfrac{145}{16}}$
On further solving the above equation, we get
$\Rightarrow \left( x+\left( \dfrac{3}{4} \right) \right)=\pm \dfrac{\sqrt{145}}{4}$
Now, we will split (+) and (-) sign to get two new equations separately, that is
$\Rightarrow \left( x+\left( \dfrac{3}{4} \right) \right)=\dfrac{\sqrt{145}}{4}$ -------- (2) and
$\Rightarrow \left( x+\left( \dfrac{3}{4} \right) \right)=-\dfrac{\sqrt{145}}{4}$ ------- (3)
Now, we will solve equation (2), which is
$\Rightarrow \left( x+\left( \dfrac{3}{4} \right) \right)=\dfrac{\sqrt{145}}{4}$
So, we will subtract $\dfrac{3}{4}$ on both sides in the above equation, we get
$\Rightarrow x+\dfrac{3}{4}-\dfrac{3}{4}=\dfrac{\sqrt{145}}{4}-\dfrac{3}{4}$
As we know, the same terms with opposite signs cancel out each other, thus we get
$\Rightarrow x=\dfrac{\sqrt{145}-3}{4}$
Thus, we will solve equation (3), which is
$\Rightarrow \left( x+\left( \dfrac{3}{4} \right) \right)=-\dfrac{\sqrt{145}}{4}$
So, we will subtract $\dfrac{3}{4}$ on both sides in the above equation, we get
$\Rightarrow x+\dfrac{3}{4}-\dfrac{3}{4}=-\dfrac{\sqrt{145}}{4}-\dfrac{3}{4}$
As we know, the same terms with opposite signs cancel out each other, thus we get
$\Rightarrow x=\dfrac{-\sqrt{145}-3}{4}$
Thus, for the equation $2{{x}^{2}}+3x-17=0$, the value of x are $\dfrac{\sqrt{145}-3}{4},\dfrac{-\sqrt{145}-3}{4}$
Note: While solving this problem, do mention all the steps properly to avoid confusion and mathematical errors. Do not forget to split (+) and (-) sign to get an accurate answer. For checking your answer, substitute the value of x on the left-hand side of the equation, then the answer must be equal to 0.
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