Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
(i) $2 x^{2}-3 x+5=0$
(ii) $3 x^{2}-4 \sqrt{3} x+4=0$
(iii) $2 x^{2}-6 x+3=0$

Answer

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Algebra

NCERT Class 10 - Math

Chapter 4

Quadratic Equations

Section 4

Solution of a Quadratic Equation by Completing the Square

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Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Video Transcript

Hello. We have question number one. We're going to find the nature of the route and they fear to exist we need to find them. So for nature of the route if X. Esquire plus bx plus equal to zero, is acquired the equation and he is the discriminatory which would be square minus four A. C. Now if D. Is less than zero, no real jobs, no real roots. Okay if D. Is equal to zero. So repeated roots, repeated roots and but really but real and when D is greater than zero. So the real end distinct routes. Okay, so let us start with problem number one that is a part one. The work that's required minus three X plus 5.20 So be equal to be a square. That is nine minus 40 so minus 31 which is negative. No real does. Okay second days second part is three excess squad minus 403 x plus four equal to zero. So they will be equal to b squared minus four is a That is 48 minus 48 0. So equal roots but real. So we need to find that route. We need to find that equal. Do okay, so 34 is 12 so this good period. And as three access points, so we'll be using quality formula basically. So quiet, that formula is X equal to minus B. That is four. Route three and plus minus zero because the zero by two into three. So for Route three by six, two by three to Route three by three and it will be repeated. Thank you. Third question is two X square minus six X plus three equal to zero. So there will be a square that is 36 minus 40 C. 24 which is 12 garden zero. So Yellen distinct routes. So we'll be using quadratic formula which is minus B so minus of minus 6, 96 plus minus B, a square minus for a C by 22. So six plus minus under 12 to 3 by four. So X will be continued three plus minus. I wrote three by two. So there are two solutions X equal to three minus Group three by two, and three plays Route three. Right to Okay, thank you so much.

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