Question

The equation whose roots are greater by 1 than those of 2x² - 3x + 1 = 0 is...

A) 3x² - 5x - 2 = 0
B) 2x² - 7x + 6 = 0
C) 2x² + 5x + 7 = 0
D) 3x² + 5x - 7 = 0.

Plz I want with Solution and PLZ don't SPAM!​

Answer 1

Use the quadratic formula

Answer 2

Answer:

$let \: the \: roots \: of \: the \: given \: equation \: \\ 2x ^{2} + 3x + 5 = 0 \: be \: a \: and \: β \\ \\ then \: a + β \: = - \frac{3}{2} and \: aβ \: = \frac{5}{2} \\ let \: the \: root \: of \: the \: required \: number \: be \: a \: and \: β \\ its \: given \: that \\ a = 3a \: and \: β \: = 3β \: \\ now \: a + β = 3a + 3β \\ 3 = (a + β) \\ = 3( - \frac{3}{2} = - \frac{9}{2} \\ also \: a \: β =( 3a)(3β) = 9aβ \\ = 9( \frac{5}{2} ) = \frac{45}{2} \\ \\ hence \: required \: answer \: is \: \\ x ^{2} - \frac{ - 9}{2} x + \frac{45}{2} = 0 \\ = 2 {x}^{2} + 9x + 45 = 2</p><p></p><p></p><p>$

Step-by-step explanation:

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