The difference between two roots of the equation ${x^3} - 13{x^2} + 15x + 189 = 0$ is 2. Then find the roots of the equation.
A. -3, 7, 9
B. -3, -7, -9
C. -3, -5, 7
D. -3, -7, 9
Answer
298.5k+ views
Hint: First suppose the roots, then obtain the relation between roots and coefficient and solve to obtain the required roots.
Formula Used:
If p, q, r be the roots of equation $a{x^3} + b{x^2} + cx + d = 0$ then the relation between roots and coefficients are,
$p + q + r = - \dfrac{b}{a}$
$pq + qr + rp = \dfrac{c}{a}$
$pqr = - \dfrac{d}{a}$
Complete step by step solution:
Suppose the roots are a, a+2, b.
Then we have,
$a + a + 2 + b = 13$
$2a + b = 11$
$b = 11 - 2a - - - - (1)$
and
$a(a + 2) + (a + 2)b + ab = 15 - - - - (2)$
and
$a(a + 2)b = - 189 - - - - (3)$
Now, from (1) and (2) we have,
$a(a + 2) + (a + 2)(11 - 2a) + a(11 - 2a) = 15$
${a^2} + 2a + 11a - 2{a^2} + 22 - 4a + 11a - 2a = 15$
$ - 3{a^2} + 20a + 7 = 0$
$3{a^2} - 20a - 7 = 0$
$3{a^2} - 21a + a - 7 = 0$
$3a(a - 7) + 1(a - 7) = 0$
$(3a + 1)(a - 7) = 0$
$a = - \dfrac{1}{3},7$
But in the option, there are no fractions so we are taking a as 7.
Hence the second root is 9 and the third one is
$11 - 2 \times 7$
$11 - 14 = -3$
Option ‘A’ is correct
Note: Sometime we tend to assume the roots as a, a+2, b and substitute one by one in the given equation to obtain a relation, that is not the correct approach, we need to use the relation between roots and coefficient to obtain the roots of the equation easily.
Formula Used:
If p, q, r be the roots of equation $a{x^3} + b{x^2} + cx + d = 0$ then the relation between roots and coefficients are,
$p + q + r = - \dfrac{b}{a}$
$pq + qr + rp = \dfrac{c}{a}$
$pqr = - \dfrac{d}{a}$
Complete step by step solution:
Suppose the roots are a, a+2, b.
Then we have,
$a + a + 2 + b = 13$
$2a + b = 11$
$b = 11 - 2a - - - - (1)$
and
$a(a + 2) + (a + 2)b + ab = 15 - - - - (2)$
and
$a(a + 2)b = - 189 - - - - (3)$
Now, from (1) and (2) we have,
$a(a + 2) + (a + 2)(11 - 2a) + a(11 - 2a) = 15$
${a^2} + 2a + 11a - 2{a^2} + 22 - 4a + 11a - 2a = 15$
$ - 3{a^2} + 20a + 7 = 0$
$3{a^2} - 20a - 7 = 0$
$3{a^2} - 21a + a - 7 = 0$
$3a(a - 7) + 1(a - 7) = 0$
$(3a + 1)(a - 7) = 0$
$a = - \dfrac{1}{3},7$
But in the option, there are no fractions so we are taking a as 7.
Hence the second root is 9 and the third one is
$11 - 2 \times 7$
$11 - 14 = -3$
Option ‘A’ is correct
Note: Sometime we tend to assume the roots as a, a+2, b and substitute one by one in the given equation to obtain a relation, that is not the correct approach, we need to use the relation between roots and coefficient to obtain the roots of the equation easily.
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