sqroot(k+9) - sqroot(k) = sqroot(3)
and
sqrt(3x+4) - sqrt(2x-7) = 3
solve these two equations??
+2446 Sure, I'll be glad to solve those equations!
1) √k+9−√k=√3
Before we start, lets determine some extraneous solutions. If one takes the square root of a negative number, the square root will result in a nonreal result, which cannot be a solution to this equation.
| k+9≥0andk≥0 | Now solve the separate inequalities. |
| k≥−9andk≥0 | Combining these two inequalities together, we can condense. |
| k≥0 | Therefore, k must be greater than or equal to zero. If it is not, we already know that it is an extraneous solution. |
Now, let's actually solve the equation.
| (√k+9−√k)2=(√3)2 | Square both sides to eliminate the radicals. When squaring a binomial,(a−b)2=a2−2ab+b2. Now, expand using this rule. |
| (√k+9)2−2√k+9√k+(√k)2=3 | Now, simplify the expansion. |
| k+9−2√k+9√k+k=3 | Combine like terms (the k and the k). |
| 2k+9−2√k+9√k=3 | Subtract 9 from both sides. |
| 2k−2√k+9√k=−6 | Divide by 2 from both sides to simplify things somewhat. |
| k−√k+9√k=−3 | Subtract k from both sides. |
| −√k+9√k=−k−3 | Now, square both sides again. This time, we will completely eliminate the radical. Divide by -1 to negate all these negative signs. |
| √k+9√k=k+3 | Square both sides. This time, the radicals will be completely eliminated. |
| (√k+9√k)2=(k+3)2 | The left hand side is an entire term multiplied collectively, so it is dealt with differently. The right hand side is expanded in the exact same fashion as the first expansion. |
| (√k+9)2(√k)2=k2+2(3)(k)+32 | |
| k(k+9)=k2+6k+9 | Now, distribute the k to all the terms in the parentheses. |
| k2+9k=k2+6k+9 | The quadratic terms will cancel out here, which simplifies this process exponentially. |
| 9k=6k+9 | Subtract 6k from both sides. |
| 3k=9 | |
| k=3 | This solution fits within k≥0, so this is the only valid solution. |
+2446 Since both solutions require a lot of effort and time, I decided to split the next one into another section.
2) √3x+4−√2x−7=3
First, let's solve for any extraneous solution.
| 3x+4≥0and2x−7≥0 | Now, solve both inequalities simultaneously. |
| 3x≥−4and2x≥7 | Divide by the coefficient of the linear term. |
| x≥−43andx≥72 | Combine these two inequalities. |
| x≥72 | |
Now that we have determined the solution set that does not result in a nonreal answer, let's continue to solve.
| √3x+4−√2x−7=3 | Just like the previous problem, square both sides. | ||
| (√3x+4−√2x−7)2=32 | Now, do the expansion. | ||
| (√3x+4)2−2(√3x+4)(√2x−7)+(√2x−7)2=9 | Now, simplify the best one can. | ||
| 3x+4−2(√3x+4)(√2x−7)+2x−7=9 | Combine like terms on the left hand side. | ||
| 5x−3−2(√3x+4)(√2x−7)=9 | Add 3 to both sides. | ||
| 5x−2(√3x+4)(√2x−7)=12 | Subtract 5x from both sides. | ||
| −2(√3x+4)(√2x−7)=−5x+12 | Divide by -1. | ||
| 2(√3x+4)(√2x−7)=5x−12 | Now, square both sides again to eliminate all instances of a radical. | ||
| [2(√3x+4)(√2x−7)]2=(5x−12)2 | |||
| 22(√3x+4)2(√2x−7)2=(5x)2−2(12)(5x)−122 | Now, simplify until you get to a quadratic. | ||
| 4(3x+4)(2x−7)=25x2−120x−144 | Now, multiply the binomials together. | ||
| 4(6x2+8x−21x−28)=25x2−120x−144 | |||
| 4(6x2−13x−28)=25x2−120x−144 | Now, distribute the 4 to all terms. | ||
| 24x2−52x−112=25x2−120x−144 | Now, move the terms to one side. | ||
| x2−68x−256=0 | Knowing that 256 is 64*4 allows one to factor this equation. | ||
| (x−64)(x−4)=0 | Now, solve for x by setting each factor equal to 0. | ||
| |||
| Both of these solutions fit within the original boundaries set by possible extraneous solutions, so both of these are solutions. | ||
+2446 Both parts took me approximately the same amount of time (30 minutes).
