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Home » Find the values of k for each of the following quadratic equations, so that they have two equal roots. (i) 2×2 + kx + 3 = 0 (ii) kx (x – 2) + 6 = 0
Find the values of k for each of the following quadratic equations, so that they have two equal roots. (i) 2×2 + kx + 3 = 0 (ii) kx (x – 2) + 6 = 0
CBSE | Class 10 | Exercise 4.4 | Quadratic Equations
Find the values of k for each of the following quadratic equations, so that they have two equal roots. (i) 2×2 + kx + 3 = 0 (ii) kx (x – 2) + 6 = 0

Solutions:

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

Comparing the given equation with a{{x}^{2}}~+~bx~+~c~=\text{ }0 , we get,

a~=\text{ }2,~b~=\text{ }k\text{ }and~c~=\text{ }3

As we know, Discriminant =~{{b}^{2}}~\text{ }4ac

=\text{ }{{\left( k \right)}^{2}}~\text{ }4\left( 2 \right)\text{ }\left( 3 \right)

=~{{k}^{2}}~\text{ }24

For equal roots, we know,

Discriminant = 0

{{k}^{2}}~\text{ }24\text{ }=\text{ }0

{{k}^{2}}~=\text{ }24

k\text{ }=\text{ }\pm \surd 24~=\text{ }\pm 2\surd 6

 

\left( ii \right)~kx\left( x~\text{ }2 \right)\text{ }+\text{ }6\text{ }=\text{ }0

or~k{{x}^{2}}~\text{ }2kx~+\text{ }6\text{ }=\text{ }0

Comparing the given equation with a{{x}^{2}}~+~bx~+~c~=\text{ }0 , we get

a~=~k,~b~=\text{ }\text{ }2k~and~c~=\text{ }6

We know, Discriminant =~{{b}^{2}}~\text{ }4ac

=\text{ }{{\left( \text{ }\text{ }2k \right)}^{2}}~\text{ }4\text{ }\left( k \right)\text{ }\left( 6 \right)

=\text{ }4{{k}^{2}}~\text{ }24k

For equal roots, we know,

{{b}^{2}}~\text{ }4ac~=\text{ }0

4{{k}^{2}}~\text{ }24k~=\text{ }0

4k~\left( k~\text{ }6 \right)\text{ }=\text{ }0

Either\text{ }4k~=\text{ }0\text{ }or~k~=\text{ }6\text{ }=\text{ }0

k~=\text{ }0\text{ }or~k~=\text{ }6

However, if k = 0, then the equation will not have the terms ‘x2‘ and ‘x‘.

Therefore, if this equation has two equal roots, k should be 6 only

i

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