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Home » Find the roots of the following quadratic equations, if they exist, by the method of completing the square: (i) 2×2 – 7x +3 = 0 (ii) 2×2 + x – 4 = 0
Find the roots of the following quadratic equations, if they exist, by the method of completing the square: (i) 2×2 – 7x +3 = 0 (ii) 2×2 + x – 4 = 0
CBSE | Class 10 | Exercise 4.3 | Quadratic Equations
Find the roots of the following quadratic equations, if they exist, by the method of completing the square: (i) 2×2 – 7x +3 = 0 (ii) 2×2 + x – 4 = 0

Solutions:

\left( \mathbf{i} \right)\text{ }\mathbf{2}{{\mathbf{x}}^{\mathbf{2}}}~-\mathbf{7x}~+\text{ }\mathbf{3}\text{ }=\text{ }\mathbf{0}

\Rightarrow 2{{x}^{2}}~-7x~=\text{ }3

Dividing by 2 on both sides, we get

\Rightarrow {{x}^{2}}~-7x/2\text{ }=\text{ }-3/2

\Rightarrow {{x}^{2~}}-2\text{ }\times \text{ }x\text{ }\times 7/4\text{ }=\text{ }-3/2

On adding (7/4)2 to both sides of equation, we get

\Rightarrow {{\left( x \right)}^{2}}-2\times x\times 7/4\text{ }+{{\left( 7/4 \right)}^{2}}~=\text{ }{{\left( 7/4 \right)}^{2}}-3/2

\Rightarrow {{\left( x-7/4 \right)}^{2}}~=\text{ }\left( 49/16 \right)\text{ }\text{ }\left( 3/2 \right)

\Rightarrow {{\left( x-7/4 \right)}^{2~}}=\text{ }25/16

\Rightarrow {{\left( x-7/4 \right)}^{2}}~=\text{ }\pm 5/4

\Rightarrow ~x~=\text{ }7/4\text{ }\pm \text{ }5/4

\Rightarrow ~x~=\text{ }7/4\text{ }+\text{ }5/4\text{ }or\text{ }x\text{ }=\text{ }7/4\text{ }\text{ }5/4

\Rightarrow x\text{ }=\text{ }12/4\text{ }or\text{ }x\text{ }=\text{ }2/4

\Rightarrow ~x\text{ }=\text{ }3\text{ }or\text{ }x\text{ }=\text{ }1/2

 

\left( \mathbf{ii} \right)\text{ }\mathbf{2}{{\mathbf{x}}^{\mathbf{2}}}~+~\mathbf{x}~-\mathbf{4}\text{ }=\text{ }\mathbf{0}

\Rightarrow 2{{x}^{2}}~+~x~=\text{ }4

Dividing both sides of the equation by 2, we get

\Rightarrow ~{{x}^{2}}~+x/2\text{ }=\text{ }2

Now on adding (1/4)2 to both sides of the equation, we get,

\Rightarrow {{\left( x \right)}^{2~}}+~2\text{ }\times ~x~\times \text{ }1/4\text{ }+\text{ }{{\left( 1/4 \right)}^{2}}~=\text{ }2~+\text{ }{{\left( 1/4 \right)}^{2}}

\Rightarrow {{\left( x~+\text{ }1/4 \right)}^{2}}~=\text{ }33/16

\Rightarrow ~x~+\text{ }1/4\text{ }=\text{ }\pm \text{ }\surd 33/4

\Rightarrow ~x~=\text{ }\pm \text{ }\surd 33/4\text{ }\text{ }1/4

\Rightarrow ~x~=\text{ }\pm \text{ }\surd 33-1/4

Therefore, either x~=\text{ }\surd 33-1/4\text{ }or~x~=\text{ }-\surd 33-1/4

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