CBSE BOARD X, asked by kumarsourav7412, 10 months ago

determine the value (s) of m for which the equation x^2+m(4x+m-1)+2=0 has real roots​

Answers

Answered by ambarkumar1
4

Answer:

m >= 2/3 for equation to have real roots

Explanation:

x { }^{2}  + m(4x + m - 1) + 2 = 0 \\ x {}^{2}  + 4mx + m {}^{2}  - m + 2 = 0

For real roots Discriminant, D >= 0

  \sqrt{b {}^{2} - 4ac }   \geqslant 0 \\  (4m) {}^{2} - 4(1)(m {}^{2}  - m + 2) \geqslant 0 \\ 16m {}^{2}  - 4m {}^{2}  + 4m - 8 \geqslant 0 \\ 12m {}^{2}  + 4m - 8 \geqslant 0 \\ 12m {}^{2}  + 12m - 8m - 8 \geqslant 0 \\ 12m(m + 1) - 8(m + 1) \geqslant 0 \\ (12m - 8)(m + 1) \geqslant 0

Now

12m - 8 >= 0 and m+1 >= 0

m >= 8/12 m >= -1

m >= 2/3

Hence taking common from both the equation we can say that m>= 2/3 so that equation has real roots.

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