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2 years ago

15

In the quadratic formula, what are a, b, and c for the following problem? - x ^ 2 + 2x = - 6

1 answer:

Katen [24]2 years ago

4 0

$\tt - x ^{2} + 2x = - 6$

Rewrite the equation in standard form i.e. ax²+bx+c=0

$\tt \: x ^{2} - 2x - 6 = 0$

Solve the quadratic equation using the quadratic formula, Put the values into the formula and then compare the values a,b & c

$\boxed{ \sf \: quadratic \: formula: x = \frac{ - b± \sqrt{b ^{2} - 4ac} }{2a} }$

Plug in the values and compare

$\tt \: x = \frac{ - ( - 2)± \sqrt{( - 2) ^{2} - 4 \times 1 \times ( - 6) } }{2 \times 1}$

On comparing we get,

a = 1
b = -2
c = -6

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2 years ago

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2 years ago

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Step-by-step explanation:

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2 years ago

[tex]cos {}^{4} α+sin {}^{4} α= \frac{1}{4} (3+cos4α)<br>Prove:<br>

asambeis [7]

Given:

$\cos^4 \alpha+\sin^4\alpha=\dfrac{1}{4}(3+\cos 4 \alpha)$

To prove:

The given statement.

Proof:

We have,

$\cos^4 \alpha+\sin^4\alpha=\dfrac{1}{4}(3+\cos 4 \alpha)$

$LHS=\cos^4 \alpha+\sin^4\alpha$

$LHS=(\cos^2 \alpha)^2+(\sin^2 \alpha)^2$

$LHS=(\cos^2 \alpha+\sin^2\alpha)^2-2\sin ^2\alpha\cos^2 \alpha$ $[\because a^2+b^2=(a+b)^2-2ab]$

$LHS=(1)^2-2(1-\cos^2 \alpha)\cos^2 \alpha$ $[\because \cos^2 \alpha+\sin^2\alpha=1]$

$LHS=1-2\cos^2 \alpha+2\cos^4 \alpha$

Now,

$RHS=\dfrac{1}{4}(3+\cos 4 \alpha)$

$RHS=\dfrac{1}{4}[3+(2\cos^2 2\alpha-1)]$ $[\because \cos 2\theta=2\cos^2\theta -1]$

$RHS=\dfrac{1}{4}[2+2\cos^2 2\alpha]$

$RHS=\dfrac{1}{4}[2+2(2\cos^2 \alpha-1)^2]$ $[\because \cos 2\theta=2\cos^2\theta -1]$

$RHS=\dfrac{1}{4}[2+2(4\cos^4 \alpha-4\cos \alpha+1)]$ $[\because (a-b)^2=a^2-2ab+b^2]$

$RHS=\dfrac{1}{4}[2+8\cos^4 \alpha-8\cos \alpha+2]$

$RHS=\dfrac{1}{4}[4+8\cos^4 \alpha-8\cos \alpha]$

$RHS=1+2\cos^4 \alpha-2\cos \alpha$

$RHS=1-2\cos^2 \alpha+2\cos^4 \alpha$

$LHS=RHS$

Hence proved.

8 0

3 years ago