Here x=3
so, putting 3 instead of x
so, putting 3 instead of x
Solve the quadratic equation 6x^2-13x-28=0 by factoring.
1 answer:
First step to factor the trinomial, multiply the constant -28 with the leading coefficient 6.
So, -28 * 6 = - 168.
Next step is to breakdown -168 into two multiples so that their addition will result the coefficient of w= - 13.
So, -168 = -21 * 8.
And addition of - 21 and 8 will give - 13.
So, next step is to replace -13x with -21x + 8. Therefore,
6x² - 21x + 8x - 28 =0
(6x² - 21x) + (8x - 28 )=0 Make the group of two terms.
3x (2x - 7) + 4 (2x - 7) = 0 Take out the common factor from each group.
(2x - 7)(3x + 4) = 0 Again take out the common factor (2x -7).
So, 2x - 7 =0, 3x + 4 = 0 Set both the factor equal to 0.
2x = 7 , 3x = -4
So,
Hope this helps you !
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Answer:
second option
Step-by-step explanation:
Given
+ 6x² + 5 = 0
let u = x², then
u² + 6u + 5 = 0 ← in standard form
(u + 1)(u + 5) = 0 ← in factored form
Equate each factor to zero and solve for u
u + 1 = 0 ⇒ u = - 1
u + 5 = 0 ⇒ u = - 5
Change u back into terms of x, that is
x² = - 1 ( take the square root of both sides )
x = ± = ± i ( noting that
= i ), and
x² = - 5 ( take the square root of both sides )
x = ± = ±
= ±
×
= ± i
Solutions are x = ± i and x = ± i