Question

Need help fast due today thanks

Answer 1

1

Move everything to the left by subtracting 25 from both sides:

$81x^2-25=0$

Both terms are squares: $81x^2$ is the square of $9x$, while 25 is the square of 5. So, we can apply the factoring

$a^2-b^2=(a+b)(a-b)$

To factor the expression as

$(9x+5)(9x-5)=0$

A multiplication is zero if and only if one of the factors is zero, so we have

$(9x+5)(9x-5)=0 \iff 9x+5=0 \lor 9x-5=0 \iff x=-\dfrac{5}{9} \lor x=\dfrac{5}{9}$

2

The discriminant of the quadratic expression $ax^2+bx+c$ is defined as

$\Delta = b^2-4ac$

So, you have

$\Delta = 0^2-4\cdot(-1)\cdot(-25) = -100$

3

Similarly, we have

$\Delta = 7^2-4\cdot(-3)\cdot 6 = 49+72=121$