Solve the equation by factoring 2x^2+13x+15=0?
回答 (9)
2018-01-15 3:48 am
2x^2 + 13x + 15 = 0
Look for two numbers that add to +13 and multiply to 2 * 15, i.e. 30: The numbers are 10 and 3
Rewrite the middle term as a sum of 6x and 5x:
2x^2 + 10x + 3x + 15 = 0
Factor it out:
2x(x + 5) + 3(x + 5) = 0
(x + 5)(2x + 3) = 0
x = -3/2, x = -5
Look for two numbers that add to +13 and multiply to 2 * 15, i.e. 30: The numbers are 10 and 3
Rewrite the middle term as a sum of 6x and 5x:
2x^2 + 10x + 3x + 15 = 0
Factor it out:
2x(x + 5) + 3(x + 5) = 0
(x + 5)(2x + 3) = 0
x = -3/2, x = -5
2018-01-15 7:03 am
2x^2 + 13x + 15 = 0
2x^2 + 10x + 3x + 15 = 0
2x(x + 5) + 3(x + 5) =0
(x + 5)(2x + 3) = 0
Solutions:
x = -5
x = -3/2
2x^2 + 10x + 3x + 15 = 0
2x(x + 5) + 3(x + 5) =0
(x + 5)(2x + 3) = 0
Solutions:
x = -5
x = -3/2
2018-01-15 3:37 am
If there are two linear factors (of the form ax + b),
then one must start with 2x and the other must start with x.
Since all the signs are positive, you must have
(2x + M)(x + N),
and you need two things: MN = 15, and 2N+M = 13.
Mess around a little bit, and you find out it works when
M = 3 and N = 5.
The factors are (2x + 3)(x + 5).
then one must start with 2x and the other must start with x.
Since all the signs are positive, you must have
(2x + M)(x + N),
and you need two things: MN = 15, and 2N+M = 13.
Mess around a little bit, and you find out it works when
M = 3 and N = 5.
The factors are (2x + 3)(x + 5).
2018-01-15 3:52 am
2x² + 13x + 15 = 0
quadratic formula
x = [-13 ± √(13² – 4·2·15)] / [2·2]
= [-13 ± √49] / 4
= [-13 ± 7] /4
= -1.5, -5
2x² + 13x + 15 = 2(x + 1.5)(x + 5) = (2x+3) (x+5)
quadratic formula
x = [-13 ± √(13² – 4·2·15)] / [2·2]
= [-13 ± √49] / 4
= [-13 ± 7] /4
= -1.5, -5
2x² + 13x + 15 = 2(x + 1.5)(x + 5) = (2x+3) (x+5)
2018-01-15 11:26 am
f(x)=2x^2+13x+15..[1]. f(-5) = 2*5^2 +13*(-5)+15 = 50-65+15 = 0. So (x+5) is a factor of f(x).
Then f(x) = (x+5)(2x+a) = 2x^2+(a+10)x+5a..[2]. Equating coefficients of like powers of x in
[1] & [2] gives (a+10) = 13 and 5a = 15, both of which --> a = 3. Then f(x) = (x+5)(2x+3) ^
f(x) = 0-->x = -5 or -(3/2).
Then f(x) = (x+5)(2x+a) = 2x^2+(a+10)x+5a..[2]. Equating coefficients of like powers of x in
[1] & [2] gives (a+10) = 13 and 5a = 15, both of which --> a = 3. Then f(x) = (x+5)(2x+3) ^
f(x) = 0-->x = -5 or -(3/2).
2018-01-15 9:41 am
2x² + 13x + 15 = 0
2x² + (13-d)x + dx + 15 = 0
...find d such that 2*15 = d(13-d), which is d² - 13d + 30 = 0 which is (d - 3)(d - 10) = 0. So choose d=3 or d=10 and continue...
2x² + (13-3)x + 3x + 15 = 0 OR 2x² + (13-10)x + 3x + 15 = 0
2x(x + 5)x + 3(x + 5) = 0 OR x(2x + 3) + 5(2x + 3) = 0
(2x + 3)(x + 5) = 0
2x + 3 = 0 OR x + 5 = 0
So the solutions are x = -3/2 and x = -5.
2x² + (13-d)x + dx + 15 = 0
...find d such that 2*15 = d(13-d), which is d² - 13d + 30 = 0 which is (d - 3)(d - 10) = 0. So choose d=3 or d=10 and continue...
2x² + (13-3)x + 3x + 15 = 0 OR 2x² + (13-10)x + 3x + 15 = 0
2x(x + 5)x + 3(x + 5) = 0 OR x(2x + 3) + 5(2x + 3) = 0
(2x + 3)(x + 5) = 0
2x + 3 = 0 OR x + 5 = 0
So the solutions are x = -3/2 and x = -5.
2018-01-15 6:45 am
( 2x + 3 ) ( x + 5 ) = 0
x = - 3/2 , x = - 5
x = - 3/2 , x = - 5
2018-01-15 3:35 am
Notice 2 * 1 = 2, 3 * 5 = 15, and 2 * 5 + 1 * 3 = 13
2x^2 + 13x + 15 = 0
(2x + 3)(x + 5) = 0
2x + 3 = 0 or x + 5 = 0
x = -3/2 or x = -5
2x^2 + 13x + 15 = 0
(2x + 3)(x + 5) = 0
2x + 3 = 0 or x + 5 = 0
x = -3/2 or x = -5
2018-01-15 4:02 am
(2x+3)(x+5) = 0
x = -5 or -3/2
x = -5 or -3/2
收錄日期: 2021-04-24 00:53:51
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