Question

Which are the solutions of x2 = 19x + 1? Will give Brainliest!

Answer 1

that's the answer hope it helps you

Answer 2

The roots of the given quadratic equation are 19.05 or -0.05  

Step-by-step explanation:

Let,

$f(x) = x^2 - 19x -1 = 0$

The above equation is the quadratic equation in x.

For a quadratic equation in x say $ax^2 + bx + c =0$

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 $\left(\frac{b}{2 a}\right)^{2}$

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

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

On taking LCM of $4a^2$ and a

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

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

As we know that,

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

Therefore,

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

There are two values for x here,

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

Or  

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

On bringing x to another side, we get,

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

Or

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

Thus,

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

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

Now, let us consider the given equation $f(x) = x^2 - 19x -1 = 0$

$b^{2}-4 a c=(-19 \times-19)-(4 \times 1 \times(-1))=361+4=365>0$

Roots are real

The roots are given as:

$x=-b \pm \frac{\sqrt{b^{2}-4 a c}}{2 a}=-(-19) \pm \frac{\sqrt{365}}{2(1)}$

$=\frac{19 \pm \sqrt{365}}{2}$

$=\frac{19 \pm 19.10}{2}$

$=\frac{38.10}{2} \text { or }-\frac{0.10}{2}$

$x=19.05 \text { or } -0.05$