Question

a and b are roots of a quadratic equation
x2 + 5x + d=0, while a and c are the roots of
the quadratic equation x' + 6x + 2d = 0. If
there is only one common root in the two
equations, then value of d is​

Answer 1

Given:-

• a and b are the the roots of the quadratic equation x² + 5x + d = 0

• a and c are the roots of the quadratic equation x² + 6x + 2d = 0.

• There is one common root in the two equations.

To Find:-

• The value of d.

Solution:-

In the two equations, a is the common root.

So, Let us substitute x = a

$\sf{x}^{2} + 5x + d = 0 \implies {a}^{2} + 5a + d = 0......(i)$

$\sf{x}^{2} + 6x + 2d = 0 \implies {a}^{2} + 6a + 2d = 0......(ii)$

Equation (ii) - (i):-

$\sf \implies {a}^{2} + 6a + 2d - ( {a}^{2} + 5a + d) = 0$

$\sf \implies {a}^{2} + 6a + 2d - {a}^{2} - 5a - d= 0$

$\sf \implies {a}^{2} - {a}^{2} + 6a - 5a + 2d - d= 0$

$\sf \implies a + d = 0$

$\sf \implies a = - d$

Substitute a = -d in equation (i):-

$\sf \implies {a}^{2} + 5a + d = 0$

$\sf \implies {(-d)}^{2} + 5(-d) + d = 0$

$\sf \implies {d}^{2} -5d + d = 0$

$\sf \implies {d}^{2} - 4d = 0$

$\sf \implies d(d - 4) = 0$

$\sf \implies d - 4 = 0$

$\sf \implies d = 4$

$\large\dashrightarrow\underline{\boxed{\bf \therefore The \: value \: of \: d = 4}}$

Answer 2

Answer :-

• Value of ‘d’ = 4

Given :-

• Roots of x² + 5x + d = 0 are ‘a’ and ‘b’.

• Roots of x² + 6x + 2d = 0 are ‘a’ and ‘c’.

To find :-

• The value of ‘d’.

Method of solution :-

x² + 5x + d = 0 - (i)

x² + 6x + 2d = 0 - (ii)

After substituting the common root ‘a’, we have to substarct eq (i) from eq (ii).

Then we get ‘a’ in terms of ‘d’.

So that we can substitute in any equation to get the value of ‘d’.

Solution :-

➙ a² + 5a + d = 0 - (i)

➙ a² + 6a + 2d = 0 - (ii)

Substarct eq (i) from eq (ii),

➙ (a² + 6a + 2d) - (a² + 5a +d) = 0 - 0

➙ a + d = 0

➙ a = -d

Substitute value of ‘a’ in eq (i),

➪ (-d)² + 5(-d) + d = 0

➪ d² - 5d + d = 0

➪ d² = 4d

➪ d = 4