Dear student, ax2+bx+c = 0let one root m then other root mnm+mn = -ba 1m×mn =ca mn+1 =ca m =ca 1n+1put in eq1ca 1n+1+ca nn+1 = -baaca 1n+1+aca nn+1=-ban c1n+1+acn 1n+1+b=0 Hope this information will clear your doubts If you have any more doubts just ask here on the forum and our experts will try to help out as soon as possible Regards

Question no A9

Question no A9 - - Y"' 'Y quadratic equation of the form X2 p + — and -z as- 10 in place of- 11 and the roots were found tobe and 6. Find the roots of the Find the value of the expression + 72 when x = 2 Find the value of the expression 2x3 + 2x2— 7x + 72 when x = Solve the following equation 2-2x + 2x+2 — 32 = O 2 -7m Let a, b, c be real numbers with a # O and let u, ß bethe roots of the equation ax2 + bx+ c the roots of a3x2 + abcx + c3 = O in terms of a, If a, ß are roots of x2—px q O and&a —02, ß + 2 are roots of x2—px + r = O, then prove that 16q + (r + 4— = 4p2. & If one root of the equation ax2 + bx + c = O is equal to nth power of the other root, show that If the sum of the roots of quadratic eqyation (a + 1)x2 + (2a + 3)x + (3a + 4) = O is —1 , then find the of the roots. Secti (B) : Relation between roots and coefficients ; Higher Degree Equations If a and be two real roots of the equation x3 + px2 + qx + r = O satisfying the relation + 1 rove that r2 + pr + q + I = O. If a, p, yare the roots of the equation x3 + px2 + qx + r = O, then find the value of 1 1 13 - AResonance website @ orporate Office. CG Tower. A-46 & 52, IPIA, Near City WI, Jhalawar Road, Kota (Raj \-32dQ Toll Free 1800 258 5555 CIN_ IJ80302RJ2007PTC024029

Dear student,

a x^{2} + b x + c = 0 l e t o n e r o o t m t h e n o t h e r r o o t m^{n} m + m^{n} = \frac{- b}{a} . . . 1 m \times m^{n} = \frac{c}{a} m^{n + 1} = \frac{c}{a} m = {(\frac{c}{a})}^{\frac{1}{n + 1}} p u t i n e q 1 {(\frac{c}{a})}^{\frac{1}{n + 1}} + {(\frac{c}{a})}^{\frac{n}{n + 1}} = \frac{- b}{a} a {(\frac{c}{a})}^{\frac{1}{n + 1}} + a {(\frac{c}{a})}^{\frac{n}{n + 1}} = - b {(a^{n} c)}^{\frac{1}{n + 1}} + {(a c^{n})}^{\frac{1}{n + 1}} + b = 0

Hope this information will clear your doubts. If you have any more doubts just ask here on the forum and our experts will try to help out as soon as possible.
Regards

View Full Answer