Find the parameter a for which the quadratic equation $(a + 1) x^{2} + 2 (a + 1) x + a - 2 = 0$ has two equal roots.

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Solution

Condition for equal roots:

$b^{2} = 4 a c$

Given,

$(a + 1) x^{2} + [2 (a + 1)] x + a - 2 = 0$

$[2 (a + 1)]^{2} = 4 (a + 1) (a - 2)$

$4 (a + 1)^{2} = 4 (a + 1) (a - 2)$

$a^{2} + 1 + 2 a = a^{2} - a - 2$

$2 a + a = - 2 - 1$

$3 a = - 3$

$∴ a = - 1$

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Find the parameter a for which the quadratic equation (a+1)x2+2(a+1)x+a−2=0 has two equal roots.

Condition for equal roots:b2=4acGiven,(a+1)x2+[2(a+1)]x+a−2=0[2(a+1)]2=4(a+1)(a−2)4(a+1)2=4(a+1)(a−2)a2+1+2a=a2−a−22a+a=−2−13a=−3∴a=−1

Find the parameter a for which the quadratic equation $(a + 1) x^{2} + 2 (a + 1) x + a - 2 = 0$ has two equal roots.

Condition for equal roots:

$b^{2} = 4 a c$

Given,

$(a + 1) x^{2} + [2 (a + 1)] x + a - 2 = 0$

$[2 (a + 1)]^{2} = 4 (a + 1) (a - 2)$

$4 (a + 1)^{2} = 4 (a + 1) (a - 2)$

$a^{2} + 1 + 2 a = a^{2} - a - 2$

$2 a + a = - 2 - 1$

$3 a = - 3$

$∴ a = - 1$