# If 1 is a Root of the Quadratic Equation 3 X 2 + a X − 2 = 0 and the Quadratic Equation a ( X 2 + 6 X ) − B = 0 Has Equal Roots, Find the Value of B. - Mathematics

If 1 is a root of the quadratic equation $3 x^2 + ax - 2 = 0$ and the quadratic equation $a( x^2 + 6x) - b = 0$ has equal roots, find the value of b.

#### Solution

The given quadratic equation is $3 x^2 + ax - 2 = 0$

and one root is 1.

Then, it satisfies the given equation.

$3 \left( 1 \right)^2 + a\left( 1 \right) - 2 = 0$

$\Rightarrow 3 + a - 2 = 0$

$\Rightarrow 1 + a = 0$

$\Rightarrow a = - 1$

The quadratic equation $a( x^2 + 6x) - b = 0$,has equal roots.

Putting the value of a, we get

$- 1\left( x^2 + 6x \right) - b = 0$

$\Rightarrow x^2 + 6x + b = 0$

Here,

$A = 1, B = 6 \text { and } C = b$

As we know that $D = B^2 - 4AC$

Putting the values of  $A = 1, B = 6 \text { and } C = b$.

$D = \left( 6 \right)^2 - 4\left( 1 \right)\left( b \right)$

$= 36 - 4b$

The given equation will have real and equal roots, if D = 0

Thus,

$36 - 4b = 0$

$\Rightarrow 4b = 36$

$\Rightarrow b = 9$

Therefore, the value of b is 9.

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#### APPEARS IN

RD Sharma Class 10 Maths