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What are the roots of (x^2-22x+79=0) by quadratic formula?(x^2-22x+79=0) के मूल द्विघात सूत्र से क्या होंगे?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: (x=11\pm\sqrt{42})(x=11\pm\sqrt{42})
Here (D=(-22)^2-4(1)(79)=168), so (x=\frac{22\pm2\sqrt{42}}{2}=11\pm\sqrt{42}). In exams, simplify (D) correctly.यहां (D=(-22)^2-4(1)(79)=168), इसलिए (x=\frac{22\pm2\sqrt{42}}{2}=11\pm\sqrt{42}) है। परीक्षा में (D) को सही सरल करें।
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