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If one of the roots of the equation x2+ax+3a=0x^2 + ax + 3a = 0 is 11, then its other root is

Solution

✅ Correct Option: 2

Given equation: x2+ax+3a=0x^2 + ax + 3a = 0

One root is x=1x = 1


Since x=1x = 1 is a root, it satisfies the equation.

12+a(1)+3a=01^2 + a(1) + 3a = 0

1+a+3a=01 + a + 3a = 0

1+4a=01 + 4a = 0

4a=−14a = -1

a=−14a = -\dfrac{1}{4}


For the quadratic equation x2+ax+3a=0x^2 + ax + 3a = 0, the sum of roots equals −a-a.

Sum of roots =−a= -a

=−(−14)= -\left(-\dfrac{1}{4}\right)

=14= \dfrac{1}{4}


Let the other root be rr.

1+r=141 + r = \dfrac{1}{4}

r=14−1r = \dfrac{1}{4} - 1

r=14−44r = \dfrac{1}{4} - \dfrac{4}{4}

r=−34r = -\dfrac{3}{4}


Therefore, the other root is −34-\dfrac{3}{4}

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