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Question 113 of 128

Q.Solve the equation 6−4x−x2=x+4\sqrt{6-4x-x^2} = x+4.

Tamil Nadu DgeTamil Nadu HSC First Year (DGE) Board 2023Subjective· 3mImportance★★★★★
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Squaring both sides of 6−4x−x2=x+4\sqrt{6-4x-x^2}=x+4 gives a quadratic with roots x=−1x=-1 and x=−5x=-5; only x=−1x=-1 satisfies the original equation.

For the square root to be valid, we need the right side x+4≥0x+4\ge0, i.e. x≥−4x\ge-4.

Square both sides:

6−4x−x2=(x+4)2=x2+8x+166-4x-x^2 = (x+4)^2 = x^2+8x+16

6−4x−x2−x2−8x−16=0  ⟹  −2x2−12x−10=06-4x-x^2-x^2-8x-16 = 0 \implies -2x^2-12x-10=0

Divide by −2-2:

x2+6x+5=0  ⟹  (x+1)(x+5)=0  ⟹  x=−1 or x=−5x^2+6x+5=0 \implies (x+1)(x+5)=0 \implies x=-1 \text{ or } x=-5

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