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CBSE Class 10 Basic Math 2026 All Sets Solved Paper

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Question : 7 of 20
Marks: +1, -0
The value of k for which the equation 2kx2−6x+3=02kx^2 - 6x + 3 = 0 has real and equal roots, is:
Solution:  
Given quadratic equation is 2kx2−6x+3=02kx^2 - 6x + 3 = 0
For a quadratic equation to have real and equal roots,
b2−4ac=0b^2 - 4ac = 0
Comparing the given equation with general equation ax2+bx+c=0ax^2 + bx + c = 0
We get a = 2k, b = −6 and c = 3
b2−4ac=0b^2 - 4ac = 0
(−6)2−4×2k×3=0(-6)^2 -4 \times 2k \times 3 = 0
36−24k=036 - 24k = 0
24k=3624k = 36
k=  3624k = \;\frac{36}{24}
∴k=  32\therefore k = \;\frac{3}{2}
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