Question

t2-3t +20=0

please solve this??​

Answer 1

${\tt{\red{\underline{\large{Given}}}}}$

• A polynomial
• t²-3t+20

${\sf{\green{\underline{\large{To\:Find}}}}}$

• Factors of the polynomial
• Relationship between cofficient

${\sf{\pink{\underline{\Large{Explanation}}}}}$

• t²-3t+20

Here,

a=1

b=-3

C=+20

From quadratic polynomial

$\tt{X=\dfrac{-b\pm{\sqrt{b^2-4ac}}}{2a}}$

$\rightarrow\tt{X=\dfrac{-b\pm{\sqrt{b^2-4ac}}}{2a}}$

$\rightarrow\tt{X=\dfrac{3\pm{\sqrt9-80}}{21}}$

• Here the value b²-4ac is <0
• Therefore they have no real roots

☞

Additional Information

$\rightarrow\tt\alpha{+}\beta{=}\dfrac{-3}{1}$

$\rightarrow\tt\alpha{+}\beta{=}\dfrac{Cofficient\:of\:X}{Cofficient\:of\:x^2}=$

&

$\rightarrow\tt\alpha{\times}\beta{=}\dfrac{20}{1}$

$\rightarrow\tt{\large\alpha{\times}\beta{=}\dfrac{Constant\:term}{Cofficient\:of\:x^2}}$

Hence,relation verified

Answer 2

Answer:

t= (3+i✓71)/2 & (3-i✓71)/2

Step-by-step explanation:

Given that t²-3t+20=0

compare the general equation ax²+bx+c=0

so, a=1, b=(-3) c=20

According to shridhracgarya formula

t=[(-b)±√(b²-4ac)]/2a

t=[(3)±✓{(-3)²-4*1*20}]/2*1

t=[3±✓(9-80)]/2

t=[3±✓(-71)]/2

t=[3±i(✓71)]/2

from positive side t=(3+i✓71)/2

from negative side t=(3-i✓71)/2

so value of t= (3+i✓71)/2 & (3-i✓71)/2