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17. If f(x) = ax^2 + bx + c has no real zeros and a +b+c0(c)c |
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Answer» JOIN oa 1. Pa = (pn-an) , PB = (pn+bn ) pa.Pb = (pn-an)(pn+bn) on perpendicular ab so an = bn pa.Pb = (pn-an)(pn+an) = pn^2-an^2 2. Pn^2-an^2 = in ∆ona oa^2 = on^2+an^2 an^2 = oa^2-on^2 pn^2-(oa^2-on^2)= pn^2+on^2-oa^2 in ∆ onp op^2 = on^2+pn^2 so. Op^2- oa^2 oa= ot pn^2-an^2= op^2-ot^2 3. From 1 & 2 pa.Pb = op^2-ot^2 in ∆ otp op^2 = pt^2+ot^2 op^2-ot^2= pt^2 so pa.Pb = pt^2 1. Pa = (pn-an) , pb = (pn+bn ) pa.Pb = (pn-an)(pn+bn) on perpendicular ab so an = bn pa.Pb = (pn-an)(pn+an) = pn^2-an^2 2. Pn^2-an^2 = in ∆ona oa^2 = on^2+an^2 an^2 = oa^2-on^2 pn^2-(oa^2-on^2)= pn^2+on^2-oa^2 in ∆ onp op^2 = on^2+pn^2 so. Op^2- oa^2 oa= ot pn^2-an^2= op^2-ot^2 3. From 1 & 2 pa.Pb = op^2-ot^2 in ∆ otp op^2 = pt^2+ot^2 op^2-ot^2= pt^2 so pa.Pb = pt^2 log in to add aJoin oa 1. Pa = (pn-an) , pb = (pn+bn ) pa.Pb = (pn-an)(pn+bn) on perpendicular ab so an = bn pa.Pb = (pn-an)(pn+an) = pn^2-an^2 2. Pn^2-an^2 = in ∆ona oa^2 = on^2+an^2 an^2 = oa^2-on^2 pn^2-(oa^2-on^2)= pn^2+on^2-oa^2 in ∆ onp op^2 = on^2+pn^2 so. Op^2- oa^2 oa= ot pn^2-an^2= op^2-ot^2 3. From 1 & 2 pa.Pb = op^2-ot^2 in ∆ otp op^2 = pt^2+ot^2 op^2-ot^2= pt^2 so pa.Pb = pt^2 log in to add a Explanation: Join oa 1. Pa = (pn-an) , pb = (pn+bn ) pa.Pb = (pn-an)(pn+bn) on perpendicular ab so an = bn pa.Pb = (pn-an)(pn+an) = pn^2-an^2 2. Pn^2-an^2 = in ∆ona oa^2 = on^2+an^2 an^2 = oa^2-on^2 pn^2-(oa^2-on^2)= pn^2+on^2-oa^2 in ∆ onp op^2 = on^2+pn^2 so. Op^2- oa^2 oa= ot pn^2-an^2= op^2-ot^2 3. From 1 & 2 pa.Pb = op^2-ot^2 in ∆ otp op^2 = pt^2+ot^2 op^2-ot^2= pt^2 so pa.Pb = pt^2 log in to add aJoin oa 1. Pa = (pn-an) , pb = (pn+bn ) pa.Pb = (pn-an)(pn+bn) on perpendicular ab so an = bn pa.Pb = (pn-an)(pn+an) = pn^2-an^2 2. Pn^2-an^2 = in ∆ona oa^2 = on^2+an^2 an^2 = oa^2-on^2 pn^2-(oa^2-on^2)= pn^2+on^2-oa^2 in ∆ onp op^2 = on^2+pn^2 so. Op^2- oa^2 oa= ot pn^2-an^2= op^2-ot^2 3. From 1 & 2 pa.Pb = op^2-ot^2 in ∆ otp op^2 = pt^2+ot^2 op^2-ot^2= pt^2 so pa.Pb = pt^2 log in to add a |
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