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Mathematics

1 answer:

REY [17]2 years ago

8 0

Answer: 45/

Step-by-step explanation:

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3x^2 + 2x + 5 = 0

Lyrx [107]

Answer:

Step-by-step explanation:

3x2−2x−5=0

Let us factorize using the split the middle term method.

On inspection we observe that middle term can be split in two parts whose product is equal to the product of first and third term.

Two terms are −5xand3x

The equation becomes

3x2+3x−5x−5=0, pairing the two and taking out common factors

⇒(3x2+3x)−(5x+5)=0,

don't forget to change the sign of −5 once placed inside the parenthesis.

⇒3x(x+1)−5(x+1)=0

⇒(x+1)(3x−5)=0

Either (x+1)=0 or (3x−5)=0

We obtain x=−1,53

or x=−1,123

3 0

2 years ago

Verify the identity (tan x + 1)^2 + (tan x-1)^2= 2 sec^2 x

Elina [12.6K]

$(\tan x+1)^2+(\tan x-1)^2=2\sec^2x\\\\\text{use}\ \tan x=\dfrac{\sin x}{\cos x}\\\\L_s=\left(\dfrac{\sin x}{\cos x}+\dfrac{\cos x}{\cos x}\right)^2+\left(\dfrac{\sin x}{\cos x}-\dfrac{\cos x}{\cos x}\right)^2\\\\=\left(\dfrac{\sin x+\cos x}{\cos x}\right)^2+\left(\dfrac{\sin x-\cos x}{\cos x}\right)^2\\\\=\dfrac{(\sin x+\cos x)^2}{\cos^2x}+\dfrac{(\sin x-\cos x)^2}{\cos^2x}\\\\\text{use}\ (a\pm b)^2=a^2\pm2ab+b^2$

$=\dfrac{\sin^2x+2\sin x\cos x+\cos^2}{\cos^2x}+\dfrac{\sin^2x-2\sin x\cos x+\cos^2}{\cos^2x}\\\\=\dfrac{\sin^2x+2\sin x\cos x+\cos^2+\sin^2x-2\sin x\cos x+\cos^2}{\cos^2x}\\\\=\dfrac{2\sin^2x+2\cos^2x}{\cos^2x}=\dfrac{2(\sin^2x+\cos^2x)}{\cos^2x}\\\\\text{use}\ \sin^2x+\cos^2x=1\\\\=\dfrac{2(1)}{\cos^2x}=2\cdot\dfrac{1}{\cos^2x}=2\left(\dfrac{1}{\cos x}\right)^2\\\\\text{use}\ \sec x=\dfrac{1}{\cos x}\\\\=2(\sec^2x)=2\sec^2x=R_s\\\\L_s=R_s\Rightarrow The\ identity$