Shweta and Sandhya solved a quadratic equation. In solving it, Shweta made a mistake in the constant

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09 Nov 2015 Equation

Shweta and Sandhya solved a quadratic equation. In solving it, Shweta made a mistake in the constant term and got the roots as 6 and 2, while Sandhya made a mistake in the coefficient of x only and obtained the roots as -7 and -1. The correct root of the equation are.

OPtion

1) 6, -1
2) -7, 2
3) -6, -2
4) 7, 1
5) 7, -2
6) 7, 6
7) -7, -6
8) 3, 2
9) 6, 2
10)None of these

Solution

When there is no mistake in a and b, the sum of roots must be correct. When there is no mistake in a and c, product of the roots must be correct.
Therefore sum of roots = 6+2 = 8 and product of roots -7*-1 = 7
So the correct equation is x^2-8x+7
(x-7) (x-1)
The roots are 7, 1

Option 4)

Correct Option: 4 Best Solution (13)

RAKESH   9 years AGO

correct eqn is

x^2 - 8x + 7 = 0
roots are 7,1

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harsha   9 years AGO

shweta roots are 6 and 2 so the equation will be
(x-6)(x-2)=0
=> x^2 -8x+12=0 (mistake is only constant term i.e 12)

Sandhya roots are -7 and -1
(x+7)(X+1)=0
=> x^2+8x+7=0 (mistake is only X term i.e +8, from shweta equation the correction will be -8)
so the corrected equation is
X^2-8x+7=0
X^2-x-7x+7=0
(X-7)(X-1)=0

so the correct roots are 7 and 1

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Arifa   9 years AGO

By using the roots 6,2 equation will be x^2-8x+12.here const.12 is wrong.by nxt roots -1.-7 eqn x^2+8x+7.here coef.of x is wrong. From these 2 eqn, correct eqn x^2-8x+7. Then roots will be 7,1. Ans option 4.7,1

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Manikandan.L   9 years AGO

Let the correct equation be:
ax2+bx+c=0

For sandya b is wrong but a and c are correct. Thus, the product of roots is the same as that of the correct equation.
roots are -7 & -1...

x1x2=c/a

-7(-1)=c/a
c/a=7

For shweta, c is wrong but a and b are correct. Thus, the sum of roots is the same as that of the correct equation.
roots are 6 & 2....

x1+x2=−b/a

6+2=−b/a

b/a=8

From the correct equation
ax^2+bx+c=0

x^2+b/ax+c/a=0

x^2-8x+7=0 → the correct equation

Solving for the correct roots

x^2−8x+7=0

(x−7)(x-1)=0

x=7 & 1

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RAHUL KUMAR   9 years AGO

sweta's eq are X^2-8X+12=0
and Sandhya's eq are X^2+8X+7=0
The correct eq are X^2-8X+7=0
so the roots of the correct eq are 7 and 1.
option (4) is correct answer.

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Kethineni Vinodh   9 years AGO

1eq=>x^2-8x+12
2eq=>x^2+8x+7
then required sum of roots=8 and product of roots=7
by option verification=7,1

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JENSI KARUNYA.S   9 years AGO

swetha got the roots=6,2 then the equation is x^2+8x+12=0(constant term is wrong)
sandhiya got roots=-1,-7 then equation is x^2-8x+7=0(co-efficient of x is wrong)
therefore the correct equn is=x^2+8x+7=0
the roots are 7,1
correct answer is option 4) 7,1

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Devendra Marghade   9 years AGO

Shweta's equation will be (x-6)(x-2)=x^2-8x+12=0 ---(i)
Sandhya's equation will be (x+7)(x+1)=x^2+8x+7=0 ---(ii)
From (i), 12 is incorrect and from (ii), 8 is incorrect
So correct equation will be x^2-8x+7=0, (x-7)(x-1)=0
So roots are 7 and 1
Option 4)

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yalamanda   9 years AGO

(x-6)(x-2)=x2-8x+12=0;(x+7)(x+1)=x2+8x+7=0
so from the two statements the final equation is x2-8x+7=0;
(x-7)(x-1)=0;x=7,x=1;

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Nandkishor jaiswal   9 years AGO

let the equation is a.x^2+b.x+c=0
sweta doing mistake so c is wrong and a,b then actual value of c is -1*-7=7
sandhya doing mistake so b is wrong and a,c is right then actual value of b -(2+6)=-8
now actual equation is
x^2+b.x+c=0
or, x^2-8.x+7=0
so actual root will be 7 and 1

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velmala gangadhar   9 years AGO

lets take a quadratic equations s^2+bs+c....
swetha answer is s^2-8s+12..
and sandhya answer is s^2+8s+7...
from given condition is ans is s^2-8s+7..
roots are 7,1.......

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Venkateswarlu   9 years AGO

for roots 6 & 2 quadratic equation=>x^2-8x+12=0
for roots -7 & -1 quadratic equation=>x^2+8x+7;
In first equation constant term is wrong and in second equation x coeffficient is wrong,,so final equation=>x^2-8x+7=0;
Roots =7,1

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RANJAN FADIA   9 years AGO

shweta equ. will be ax^2-8ax+12a =0
frm root.

similarly sandhya eqn will be
ax^2+8ax +7a =0


now correct eqn will ax^2-8ax +7a =0

so root is 7 & 1 ans.

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