If both the roots of the quadratic equation A(8x2 + 4) + Bx + 4x2 − 8 = 0 and 3A(2x2 + 1) + Kx + 3x2 − 6 = 0 are common, then what is the value of 9B2 − 16K2?

Correct Option: Option 4: 0

Step-by-Step Explanation:

Step 1: Simplify and express both quadratic equations in standard form ($a{x}^2+bx+c=0$).

Let the first quadratic equation be:

$A(8x^2+4)+Bx+4x^2-8=0$

Expanding and grouping like terms:

$8A{x}^2+4A+Bx+4x^2-8=0$

$(8A+4)x^2+Bx+(4A-8)=0$    --- (Equation 1)

Now, let the second quadratic equation be:

$3A(2x^2+1)+Kx+3x^2-6=0$

Expanding and grouping like terms:

$6A{x}^2+3A+Kx+3x^2-6=0$

$(6A+3)x^2+Kx+(3A-6)=0$    --- (Equation 2)

Step 2: Apply the condition for both roots to be common.

For two quadratic equations $a_1{x}^2+b_{1}x+c_1=0$ and $a_2{x}^2+b_{2}x+c_2=0$ to have both roots common, their corresponding coefficients must be proportional:

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

Substituting the coefficients from Equation 1 and Equation 2:

$\frac{8A+4}{6A+3}=\frac{B}{K}=\frac{4A-8}{3A-6}$

Step 3: Simplify the ratios of the coefficients.

Notice that we can factor out common terms in the first ratio:

$\frac{8A+4}{6A+3}=\frac{4(2A+1)}{3(2A+1)}=\frac{4}{3}$

Similarly, factor out common terms in the third ratio:

$\frac{4A-8}{3A-6}=\frac{4(A-2)}{3(A-2)}=\frac{4}{3}$

Therefore, we get:

$\frac{B}{K}=\frac{4}{3}$

Cross-multiplying gives:

$3B=4K$

Step 4: Calculate the value of 9B² − 16K².

Squaring both sides of $3B=4K$:

$(3B{)}^2=(4K{)}^2$

$9B^2=16K^2$

Subtracting $16K^2$ from both sides:

$9B^2-16K^2=0$

Thus, the value of 9B² − 16K² is 0.