Algebra · real student question

Solve (x^2 - 5x) * sqrt(4 + 3x - x^2) = 0 and give the sum of its solutions.

Question

Solve (x25x)4+3xx2=0\left(x^{2}-5x\right)\sqrt{4+3x-x^{2}}=0 and state the sum of its solutions.

Step-by-step solution

  1. Find the domain before solving anything. The radicand must be non-negative: 4+3xx204+3x-x^{2}\ge 0, i.e. x23x40x^{2}-3x-4\le 0, i.e. (x4)(x+1)0(x-4)(x+1)\le 0, so x[1,4].x\in[-1,4]. Any candidate outside this interval is not a solution of the original equation, no matter what the algebra says.

  2. Apply the zero-product rule. A product is zero exactly when one factor is zero, so either x25x=0x^{2}-5x=0 or 4+3xx2=0\sqrt{4+3x-x^{2}}=0.

  3. Solve the polynomial factor and filter by the domain. x25x=x(x5)=0x^{2}-5x=x(x-5)=0 gives x=0x=0 or x=5x=5. Since 5[1,4]5\notin[-1,4], only x=0x=0 survives; at x=5x=5 the radicand would be 4+1525=6<04+15-25=-6<0.

  4. Solve the radical factor. A square root is zero exactly when its radicand is zero: 4+3xx2=04+3x-x^{2}=0, i.e. x23x4=0x^{2}-3x-4=0, i.e. (x4)(x+1)=0(x-4)(x+1)=0, giving x=4x=4 and x=1x=-1. Both are the endpoints of the domain, so both are admissible.

  5. Collect and add. The solution set is {1,0,4}\{-1,0,4\} and the required sum is 1+0+4=3-1+0+4=3. Substituting each value back gives 000\cdot\sqrt{0}, 020\cdot 2 and 40-4\cdot 0, all zero, so all three are genuine.

Answer

x{1,0,4},sum=3x\in\{-1,\,0,\,4\},\qquad \text{sum}=3

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