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Find the roots of the following quadratic equations (if they exist) by the method of completing the square. 3. 3{{x}^{2}}+11x+10=0
Class 10 | ICSE | Maths | Quadratic Equations | RD Sharma
Find the roots of the following quadratic equations (if they exist) by the method of completing the square. 3. 3{{x}^{2}}+11x+10=0

Quadratic is a type of problem which deals with a variable multiplied by itself- an operation also known as squaring.

Solution:

Given equation,

{{x}^{2}}+\frac{11}{3}+\frac{10}{3}=0

{{x}^{2}}+2\times \frac{1}{2}\times \frac{11x}{3}+\frac{10}{3}=0

{{x}^{2}}+2\times \frac{11x}{6}+{{\left( \frac{11}{6} \right)}^{2}}-{{\left( \frac{11}{6} \right)}^{2}}+\frac{10}{3}=0

{{\left( x+\frac{11}{6} \right)}^{2}}={{\left( \frac{11}{6} \right)}^{2}}-\frac{10}{3}

{{\left( x+\frac{11}{6} \right)}^{2}}=\frac{121}{36}-\frac{10}{3}

{{\left( x+\frac{11}{6} \right)}^{2}}=\frac{121-120}{36}

{{\left( x+\frac{11}{6} \right)}^{2}}=\frac{1}{36}

{{\left( x+\frac{11}{6} \right)}^{2}}={{\left( \frac{1}{6} \right)}^{2}}

x+\frac{11}{6}=\pm \frac{1}{6}

x+\frac{11}{6}=\frac{1}{6}orx+\frac{11}{6}=\frac{-1}{6}

x=\frac{1}{6}-\frac{11}{6}orx=\frac{-1}{6}-\frac{11}{6}

x=\frac{-10}{6}orx=\frac{-12}{6}=-2

{{x}^{2}}+\frac{11x}{3}+\frac{10}{3}=0

{{x}^{2}}+2\times \frac{1}{2}\times \frac{11x}{3}+\frac{10}{3}=0

{{x}^{2}}+2\times \frac{11x}{6}+{{\left( \frac{11}{6} \right)}^{2}}-{{\left( \frac{11}{6} \right)}^{2}}+\frac{10}{3}=0

{{\left( x+\frac{11}{6} \right)}^{2}}={{\left( \frac{11}{6} \right)}^{2}}-\frac{10}{3}

{{\left( x+\frac{11}{6} \right)}^{2}}=\frac{121}{36}-\frac{10}{3}

{{\left( x+\frac{11}{6} \right)}^{2}}=\frac{1}{36}

{{\left( x+\frac{11}{6} \right)}^{2}}={{\left( \frac{1}{6} \right)}^{2}}

x+\frac{11}{6}=\pm \frac{1}{6}

x+\frac{11}{6}=\frac{1}{6}orx+\frac{11}{6}=\frac{-1}{6}

x=\frac{1}{6}-\frac{11}{6}orx=\frac{-1}{6}-\frac{11}{6}

x=\frac{-10}{6}oex=\frac{-12}{6}=-2

⇒ x=-5/3=3 or x=-2

Thus, the roots of the given quadratic equation are x=-5/3 and x=-2.

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