LT15 Abby and Robert are each trying to solve the equation x2−10x+26=0. They know that the solutions to x2=−1 are i and −i, but they are not sure how to use this information to solve for x in their equation. Solve the equation and explain to them where the i is needed.
Q. LT15 Abby and Robert are each trying to solve the equation x2−10x+26=0. They know that the solutions to x2=−1 are i and −i, but they are not sure how to use this information to solve for x in their equation. Solve the equation and explain to them where the i is needed.
Rewrite Equation: Step 1: Rewrite the equation to isolate the square root term.x2−10x+26=0⇒10x+26=x2
Square Both Sides: Step 2: Square both sides to eliminate the square root.(10x+26)2=(x2)2⇒10x+26=x4
Rearrange Polynomial: Step 3: Rearrange the equation to form a standard polynomial equation. x4−10x−26=0
Check Rational Roots: Step 4: Check for possible rational roots using the Rational Root Theorem.Possible rational roots could be factors of −26 divided by factors of 1 (leading coefficient).Possible roots: ±1, ±2, ±13, ±26
Test Possible Roots: Step 5: Test the possible roots in the polynomial.x=1: 14−10⋅1−26=1−10−26=−35 (not a root)x=−1: (−1)4−10⋅(−1)−26=1+10−26=−15 (not a root)x=2: 24−10⋅2−26=16−20−26=−30 (not a root)x=−2: (−2)4−10⋅(−2)−26=16+20−26=10 (not a root)x=13: 134−10⋅13−26=28561−130−26=28405 (not a root)14−10⋅1−26=1−10−26=−350: 14−10⋅1−26=1−10−26=−351 (not a root)14−10⋅1−26=1−10−26=−352: 14−10⋅1−26=1−10−26=−353 (not a root)14−10⋅1−26=1−10−26=−354: 14−10⋅1−26=1−10−26=−355 (not a root)
Consider Complex Roots: Step 6: Since no rational roots are found, consider complex roots or use numerical methods or graphing to find roots.This step involves higher-level mathematics or graphing calculators, which might not be readily available.