Question

The formula d=b2−4ac is used to calculate the discriminant.

Solve for b.




b=±√2ac−d

b=±√4ac−d

​ b=±√d−4ac

b=±√d+4ac

Answer 1

b=±√(d+4ac)...........

Answer 2

$\bold{b=\pm \sqrt{4 a c+d}}$

Step-by-step explanation:

Let us define $f(x)=a x^{2}+b x+c=0$ be the quadratic polynomial

For finding the roots of the equation f(x) = 0, let us expand the equation as follows:

$a x^{2}+b x+c=0$

Divide the whole equation by a

$x^{2}+\frac{b x}{a}+\frac{c}{a}=0$

To convert the above in the form of $(a+b)^{2}$, add and subtract $\left(\frac{b}{2 a}\right)^{2}$

$x^{2}+\frac{2 x b}{2 a}+\left(\frac{b}{2 a}\right)^{2}-\left(\frac{b}{2 a}\right)^{2}+\frac{c}{a}=0$

$\left(x+\left(\frac{b}{2 a}\right)\right)^{2}-\frac{b^{2}}{4 a^{2}}+\frac{c}{a}=0$

Taking LCM of $4a^{2}$ and a

$\left(x+\left(\frac{b}{2 a}\right)\right)^{2}-\frac{b^{2}-4 a c}{4 a^{2}}=0$

$\left(x+\left(\frac{b}{2 a}\right)\right)^{2}-\sqrt{\frac{b^{2}-4 a c}{4 a^{2}}}=0$

$\because\left(a^{2}-b^{2}\right)=(a+b)(a-b)$

$\left(x+\left(\frac{b}{2 a}\right)-\frac{\sqrt{b^{2}-4 a c}}{2 a}\right)\left(x+\left(\frac{b}{2 a}\right)+\frac{\sqrt{b^{2}-4 a c}}{2 a}\right)=0$

There are two values for x here.

$\left(x+\left(\frac{b}{2 a}\right)-\frac{\sqrt{b^{2}-4 a c}}{2 a}\right)=0 \text { or }\left(x+\left(\frac{b}{2 a}\right)+\frac{\sqrt{b^{2}-4 a c}}{2 a}\right)=0$

$x=-\left(\left(\frac{b}{2 a}\right)+\frac{\sqrt{b^{2}-4 a c}}{2 a}\right) \text { or } x=-\left(\left(\frac{b}{2 a}\right)-\frac{\sqrt{b^{2}-4 a c}}{2 a}\right)$

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

The above implies that $\sqrt{b^{2}-4 a c}$ is the discriminant which is the deciding factor for real, imaginary or equal roots.

If discriminant is = 0, roots are equal

If discriminant is > 0, roots are real

If discriminant is < 0, roots are imaginary

The given formula is $d=b^{2}-4 a c$

$b^{2}-4 a c-d=0$ is a quadratic polynomial in b with no  

For the above equation to be having real roots,

Discriminant > 0

Here,

$\left(b^{2}-4 a c\right)=0-(4 \times 1 \times(-d-4 a c))=4 d+16 a c d$

$b^{2}-4 a c-d=0$

Here there is only $b^2$ term, no b term available,  

$b^{2}=4 a c+d$

Taking square root on both sides

$b=\pm \sqrt{4 a c+d}$