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Question 104 of 128

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

(b) Prove that in any △ABC\triangle ABC, Δ=s(s−a)(s−b)(s−c)\Delta=\sqrt{s(s-a)(s-b)(s-c)}, where ss is the semi-perimeter of △ABC\triangle ABC.
Puducherry TnboardTamil Nadu HSC First Year (DGE) Board 2019Subjective· 5mImportance★★★★★
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Squaring 6−4x−x2=x+4\sqrt{6-4x-x^2}=x+4 gives a quadratic with roots x=−1x=-1 and x=−5x=-5; substituting back into the original equation shows only x=−1x=-1 works (the other is extraneous).

Since the left side 6−4x−x2≥0\sqrt{6-4x-x^2}\ge 0, we need x+4≥0x+4\ge 0, i.e. x≥−4x\ge -4, for the equation to have a chance of holding.

Square both sides: 6−4x−x2=(x+4)2=x2+8x+166-4x-x^2 = (x+4)^2 = x^2+8x+16.

Bring everything to one side: 6−4x−x2−x2−8x−16=0⇒−2x2−12x−10=06-4x-x^2 - x^2-8x-16 = 0 \Rightarrow -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 \Rightarrow (x+1)(x+5)=0 \Rightarrow x=-1 \text{ or } x=-5.

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