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Q.If x2+2(k+2)x+9k=0x^2+2(k+2)x+9k=0 has equal roots, find k.

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2026Subjective· 2mImportance★★★★★
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Sets the discriminant of the quadratic to zero (the condition for equal roots) and solves the resulting quadratic in kk.

  1. For x2+2(k+2)x+9k=0x^2+2(k+2)x+9k=0, compare with ax2+bx+c=0ax^2+bx+c=0: a=1, b=2(k+2), c=9ka=1,\ b=2(k+2),\ c=9k.
  2. Roots are equal when the discriminant b2−4ac=0b^2-4ac=0: [2(k+2)]2−4(1)(9k)=0[2(k+2)]^2-4(1)(9k)=0.
  3. 4(k+2)2−36k=04(k+2)^2-36k=0. Divide by 44: (k+2)2−9k=0(k+2)^2-9k=0.
  4. Expand: k2+4k+4−9k=0⇒k2−5k+4=0k^2+4k+4-9k=0\Rightarrow k^2-5k+4=0. …

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