why we multiply a*1/a in eq 3x^2+11x+k=0 so value of k we have to find so I'm confused
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solve the equation:
x2+(k−3)x+k=0⟹x=3−k±k2−6k+9−4⋅1⋅k−−−−−−−−−−−−−−−−−√2⟹x=3−k±k2−10k+9−−−−−−−−−−√2x2+(k−3)x+k=0⟹x=3−k±k2−6k+9−4⋅1⋅k2⟹x=3−k±k2−10k+92
For starters, we must have k2−10k+9=(k−1)(k−9)≥0k2−10k+9=(k−1)(k−9)≥0, which happens if k≤1k≤1 or k≥9k≥9. Recall that two numbers have the same sign if and only if their product is positive, and by Vieta's formulas, their product is kk. So far, we have 0<k≤10<k≤1or k≥9k≥9. However, k=1k=1 gives 11as a double root and 99 gives −3−3as a double root. So the book forgot these cases: our final answer is 0<k≤10<k≤1 or k≥9k≥9.
x2+(k−3)x+k=0⟹x=3−k±k2−6k+9−4⋅1⋅k−−−−−−−−−−−−−−−−−√2⟹x=3−k±k2−10k+9−−−−−−−−−−√2x2+(k−3)x+k=0⟹x=3−k±k2−6k+9−4⋅1⋅k2⟹x=3−k±k2−10k+92
For starters, we must have k2−10k+9=(k−1)(k−9)≥0k2−10k+9=(k−1)(k−9)≥0, which happens if k≤1k≤1 or k≥9k≥9. Recall that two numbers have the same sign if and only if their product is positive, and by Vieta's formulas, their product is kk. So far, we have 0<k≤10<k≤1or k≥9k≥9. However, k=1k=1 gives 11as a double root and 99 gives −3−3as a double root. So the book forgot these cases: our final answer is 0<k≤10<k≤1 or k≥9k≥9.
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