Q-34- If the roots of the equation ax2-bx-c-0 are of the form -945- -945-1 -160-and -160- -945-1 -945- then find the value of a-b-c2

Hi, #160;here #160;for #160;the #160;ax2+bx+c #160;= #160;0 #160;product #160;of #160;roots #160;= #160; #945; #945; 1 #215; #945; ...

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Standard XII

Mathematics

Nature of Roots

Q.34. If the ...

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Q.34. If the roots of the equation $a x^{2} + b x + c = 0$ are of the form $\frac{α}{α - 1} a n d \frac{α + 1}{α}$ then find the value of ${(a + b + c)}^{2}$

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$Hi, here for the {ax}^{2} + bx + c = 0 product of roots = \frac{α}{α - 1} \times \frac{α + 1}{α} = \frac{c}{a} \frac{α + 1}{α - 1} = \frac{c}{a} using componendo - dividendo rule \frac{α + 1 + α - 1}{α + 1 - α + 1} = \frac{c + a}{c - a} \frac{α}{1} = \frac{c + a}{c - a} or α = \frac{c + a}{c - a} So so \frac{α}{α - 1} = \frac{\frac{c + a}{c - a}}{\frac{c + a}{c - a} - 1} = \frac{\frac{c + a}{c - a}}{\frac{c + a - c + a}{c - a}} = \frac{c + a}{2 a} now since \frac{α}{α - 1} = \frac{c + a}{2 a} is a root of {ax}^{2} + bx + c = 0 so it will satisfy it a {(\frac{c + a}{2 a})}^{2} + b (\frac{c + a}{2 a}) + c = 0 \frac{{(c + a)}^{2}}{4 a} + b (\frac{c + a}{2 a}) + c = 0 {(c + a)}^{2} + 2 b (c + a) + 4 ac = 0 add b^{2} both side b^{2} + {(c + a)}^{2} + 2 b (c + a) + 4 ac = b^{2} {(b + (a + c))}^{2} + 4 ac = b^{2} [Using x^{2} + y^{2} + 2 xy = {(x + y)}^{2}] {(a + b + c)}^{2} = b^{2} - 4 ac$

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Q.34. If the roots of the equation ax2+bx+c=0 are of the form αα-1 and α+1α then find the value of a+b+c2

Q.34. If the roots of the equation $a x^{2} + b x + c = 0$ are of the form $\frac{α}{α - 1} a n d \frac{α + 1}{α}$ then find the value of ${(a + b + c)}^{2}$