Question

find the nature of Roots of given quadratic equation ,if real roots exist find them x²-x-90=0​

Answer 1

Answer:

real and distinct

Step-by-step explanation:

D=b2-4ac

D=(1)2-4*1*(-90)

D=1+360

D=361

so real and distict

Answer 2

QUESTION⤵️

find the nature of Roots of given quadratic equation ,if real roots exist find them x²-x-90=0

ANSWER⤵️

GIVEN:::-

${x}^{2} - x - 90 = 0$

comparing with the standard quadratic equation is

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

HERE, a=1, b= -1 , c= -90

NOW,

$D = {b}^{2} - 4ac$

$= {( - 1)}^{2} - 4 \times 1 \times ( - 90)$

$= 1 + 360$

$D = \: 361$

HERE, THE DISCRIMINANT IS 361 AND IS POSITIVE, SO THE ROOTS ARE REAL AND THE POSITIVE REAL ROOTS WILL BE THE VALUE OF X.

$x = \frac{ - b \: plus \: minus \sqrt{ {b}^{2} - 4ac } }{2a}$

$x = \frac{ 1 \: plus \: minus \: \sqrt{1 + 360} : }{2}$

$x = \frac{ \ 1 \: plus \: minus \: \sqrt{361} }{2}$

$x = \frac{ 1 \: plus \: minus \: 19 }{2}$

$x = \frac{1 + 19}{2} and \: x = \: \frac{1 - 19}{2}$

$x = 10 \: and \: x = -9</p><p></p><p>x=10andx=−9$

SO, THE REAL ROOTS ARE 10, -9

BE BRAINLY ♡

$\huge✝\mathtt\pink{hope \: it \: helps} ✝✝hopeithelps✝$