Question

Determine whether the given quadratic equation has real roots and if so, find the roots. =6x2 + x-2=0

Answer 1

Answer:

$answer \\ x = \frac{1}{2} \\ \\ x = \frac{ - 2}{3}$

Step-by-step explanation:

$\large \bold \red{ \underline{ \underline{to \: \: find}}} \\ quadratic \: \: equation \: \: has \: \: real \: \: roots \: \: and \: \: if \: \: so \\ find \: \: the \: \: roots \: \: of \: \: equations \: \: 6 {x}^{2} + x - 2 = 0 \\ \large \bold \orange{ \underline{ \underline{explanation}}} \\ \large \bold \green{notes} \\ 1) = \: if \: \: d > 0 \: \: roots \: \: are \: \: real \: \: and \: \: unequal \\ 2) = if \: \: d = 0 \: \: roots \: \: are \: \: real \: \: and \: \: equal \\ 3) =if \: \: d < 0 \: \: roots \: \: are \: \: imaginary \\ \\ \large \bold \pink{check \: \: the \: \: nature \: \: of \: \: roots} \\ 6 {x}^{2} + x - 2 = 0 \\ d = {b}^{2} - 4ac \\ d = {1}^{2} - 4(6)( - 2) \\ d = 1 + 48 \\ d = 49 \\ it \: \: means \: \: roots \: \: are \: \: real \: \: and \: \: equal \\ x = \frac{ - b + \sqrt{d} }{2a} \\ x = \frac{ - 1 + \sqrt{49} }{12} \\ \\ x = \frac{ - 1 + 7}{12} \\ \\ x = \frac{6}{12} \\ \\ x = \frac{1}{2} \\ \\ x = \frac{ - b - \sqrt{d} }{2a} \\ \\ x = \frac{ - 1 - \sqrt{49} }{12} \\ \\ x = \frac{ - 1 - 7}{12} \\ \\ x = \frac{ - 8}{12} \\ \\ x = \frac{ - 2}{3}$

Answer 2

QUESTION:

Determine whether the given quadratic equation has real roots and if so, find the roots. =6x2 + x-2=0

TO FIND :

the given quadratic equation has real roots and if so, find the roots. =6x2 + x-2=0

ANSWER:

We know that quadratic equation is in the form of

$\huge\red {a {x}^{2} + bx + c = 0}$

$\purple {(we \: check \: discriminant \: ( {b}^{2} - 4ac) \: to \: find \: which \: types \: of \: roots \: there \: is..)}$

$\green{If}$

1.

$\red {d greater than 0 \\ (real \: and \: unequal \: roots)}$

2.

$\blue {d = 0 \\ (real \: and \: equal \: roots)}$

3.

$\orange {d less than 0 \\ (imaginary \: roots)}$

Now come to main question ;

$6 {x}^{2} + x - 2 = 0$

here;

a = 6

b =1

c = -2.

so, using the formula;

$d = {b}^{2} - 4ac \\ d = {1}^{2} - 4 \times 6 \times ( - 2) \\ d = 1 + 48 \\ d = 49$

hence, discriminant is 49 that is greater than 1 so,

$\organe{(it \: has \: real \: and \: unequal \: roots)}$

Now to find the roots we use quadratic formula;

$x = \frac{ - 1 \binom{ + }{ - } \sqrt{ {1}^{2} - 4 \times 6 \times ( - 2)} }{2 \times 6}$

$x = \frac{ - 1 \binom{ + }{ - } \sqrt{49} }{12} \\ x = \frac{ - 1 \binom{ + }{ - } 7}{12}$

Taking (-) first;

$x = \frac{ - 1 - 7}{12} \\ x = \frac{ - 8}{12} \\ x = \frac{ - 2}{</u><u>3</u><u>}$

Taking (+);

$x = \frac{ - 1 +7 }{12} \\ x = \frac{ - 6}{12} \\ x = \frac{ - 1}{2}$

Hence roots are;

$\frac{ - 2}{3} \: and \: \frac{ - 1}{2}$