Question

Solve the quadratic equation 2xsqure -6x+7=0

Answer 1

Answer:

Step-by-step explanation:

It has imaginary roots as determinant b^2-4ac is less tham zero

Answer 2

Given:-

Solve the quadratic equation 2xsqure -6x+7

Find:-

2xsqure -6x+7=0??

Solution:-

$\sf \: Given \: as 2x ^2  – (3 + 7i)x + (9i – 3) =$

Now, we shall apply discriminant rule,

$\sf \: Where, x = { \underline{ \boxed{ \sf{ \red{(-b± √(b2 – 4ac))/2a}}}}}$

Here, a = 2, b = -(3 + 7i), c = (9i – 3)

Therefore,

$\sf{ \underline{x = - ( - (3 + 71)± \sqrt{{( - (3 + 71))^2 - 4(2)(9i - 3)}}}} \\ 2(2)$

${ \underline{ \sf = (3 + 7i)± \sqrt{(3 + 7i)^{2} - 8(9i - 3}}} \\ 4$

${ \underline{ \sf= (3 + 7i)± \sqrt{9 + 42i + 491^2 - 72i + 24 } }} \\ 4$

${ \underline{ \sf = (3 + 7i)± \sqrt{33 - 30i + 49i ^2}}} \\ 4$

$\sf \: we \: have \: i ^{2} = - 1$

$\sf \color{red}by \: substituting - 1 = i^{2}in \: the \: above \: equation \: we \: get$

${ \underline{ \sf = (3 + 7i)± \sqrt{33 - 30i + 49( - 1)} }} \\ 4$

${ \underline{ \sf= (3 + 7i)± \sqrt{33 - 30i - 49}}} \\ 4$

${ \underline{ \sf= (3 + 7i)± \sqrt{ - 16 - 30i}}} \\ 4$

${ \underline{ \sf(3 +7i)± \sqrt{( - 1)(16 + 30i)}}} \\ 4$

${ \underline{ \sf= (3 - 7i) ±\sqrt{i ^{2}(16 + 30i) }}} \\ 4$

${ \underline{ \sf= (3 + 7i)± \sqrt{16 + 30i}}} \\ 4$

Now, we can write 16 + 30i = 25 – 9 + 30i

16 + 30i = 25 + 9(–1) + 30i

$\sf= 25 + 9i {}^{2}  + 30i [∵ i {}^{2} = –1]$

$\sf= (5 + 3i) {}^{2} [∵ (a + b) {}^{2}  = a {}^{2} + b {}^{2} + 2ab]$

On using the result 16 + 30i = (5 + 3i)2, we get

${ \underline{ \sf x = (3 + 7i)± i\sqrt{(5 + 3i) ^2}}} \\ 4$

${ \underline{ \sf = (3 + 7i)±(5 + 3i)}} \\ 4$

${ \underline{ \sf= (3 + 7i) + i(5 + 3i)}} \\ 4$

or

${ \underline{ \sf(3 + 7i) + - (5 + 3i)}} \\ 4$

${ \underline{ \sf= 3 + 7i + 5i + 3i^2}} \\ 4$

or

${ \underline{ \sf= 3 + 7i - 5i - 3i ^2}} \\ 4$

${ \underline{ \sf = 3 + 12i + 3i ^{2}}} \\ 4$

or

${ \underline{ \sf3 + 2i - 3i {}^{2}}} \\ 4$

$\small{ \underline{ \sf= 3 + 12i + 3( - 1)}} \\ 4$

or

$\small{ \underline{ \sf{3 + 2i - 3( - 1)}}} \\ 4$

${ \underline{ \sf= 3 + 12i - 3}} \\ 4$

or

${ \underline{ \sf3 + 2i + 3}} \\ 4$

$\sf= \frac{12}{4}i \: or \frac{6 + 2i}{4}$

$\sf= 3i \: or \frac{6}{4} + \frac{2}{4}i$

$\sf \: x = 3i \: or \frac{3}{2} + \frac{1}{2}i$

$\sf3i \: or(3 + i)/2$

Therefore, the roots of the given equation are (3 + i)/2 , 3i