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CBSE Class 10 Math 2024 All India Set 1 Solved Paper

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Question : 17 of 20
Marks: +1, -0
The value of k for which the roots of the quadratic equation 6x2+4kx+k=06x^2 + 4kx + k = 0 are real and equal, is
Solution:  
Given equation is 6x2+4kx+k=06 x^2+4 k x+k=0
On comparing with ax2+bx+c=0,a x^2+b x+c=0, we get
a = 6, b = 4k and c = k
For real and Equal roots, the discriminant must be zero.
D=b2−4ac=0D=b^2-4 a c=0
(4k)2−4×6×k=0(4 k)^2-4 \times 6 \times k=0
16k2−24k=016 k^2-24 k=0
8k(2k−3)=08 k(2 k-3)=0
8k=0 or 2k−3=08 k=0 \text{ or } 2 k-3=0
k=0 or 2k=3k=0 \text{ or } 2 k =3
k = 0 or k=  32k = \; \frac{3}{2}
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