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3 years ago

Li Juan solves the equation below by first squaring both sides of the equation.

Mathematics

2 answers:

emmasim [6.3K]3 years ago

8 0

-3w + 13
Happy New Year

ANTONII [103]3 years ago

4 0

Answer:

Step-by-step explanation:

(13-2w)^2 = 169 - 52w + 4w^2

(w+6)^2 = w^2 + 12w + 36

Setting them equal:

169 - 52w + 4w^2 = w^2 + 12w + 36

3w^2 - 64w + 133 = 0

Using the quadratic formula, we get

x = (64 +/- sqrt(64^2 - 4*3*133))/6

64^2-12*133 = 2500. sqrt(2500) = 50.

x = (64 +/- 50)/6.

x = 19 or 14/6 = 7/3.

Solving the original equation:

7 = 3w.

w = 7/3.

Therefore 19 is extraneous.

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Answer:

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Step-by-step explanation:

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The quadratic equation 8x²+12x-14 has two real roots. What is the sum of the squares of these roots?

mars1129 [50]

Answer:

The real roots are

$x=\frac{(-3+\sqrt{37})}{4}$ and $x=\frac{(-3-\sqrt{37})}{4}$

The sum of the squares of these roots is $\frac{-3}{2}$

Step-by-step explanation:

The given quadratic equation is $8x^2+12x-14$ has two real roots.

To find the roots .

$8x^2+12x-14=0$

Dividing the above equation by 2

$\frac{1}{2}(8x^2+12x-14)=\frac{0}{2}$

$4x^2+6x-7=0$

For quadratic equation $ax^2+bx+c=0$ the solution is $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Where a and b are coefficents of $x^2$ and x respectively, c is a constant.

For given quadratic equation

a=4, b=6, c=-7

$x=\frac{-6\pm\sqrt{6^2-4(4)(-7)}}{2(4)}$

$=\frac{-6\pm\sqrt{36+112}}{8}$

$=\frac{-6\pm\sqrt{148}}{8}$

$=\frac{-6\pm\sqrt{37\times 4}}{8}$

$=\frac{-6\pm\sqrt{37}\times\sqrt{4}}{8}$

$=\frac{-6\pm\sqrt{37}\times 2}{8}$

$=2\frac{(-3\pm\sqrt{37})}{8}$

$=\frac{-3\pm\sqrt{37}}{4}$

$x=\frac{(-3\pm\sqrt{37})}{4}$

The real roots are

$x=\frac{(-3+\sqrt{37})}{4}$ and $x=\frac{(-3-\sqrt{37})}{4}$

Now to find the sum of the squares of these roots

$\left[\frac{-3+\sqrt{37}}{4}+\frac{(-3-\sqrt{37})}{4}\right]^2=\frac{-3+\sqrt{37}-3-\sqrt{37}}{4}$

$=\frac{-6}{4}$

$=\frac{-3}{2}$

$\left[\frac{-3+\sqrt{37}}{4}+\frac{(-3-\sqrt{37})}{4}\right]^2=\frac{-3}{2}$

Therefore the sum of the squares of these roots is $\frac{-3}{2}$

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LUCKY_DIMON [66]

9514 1404 393

Answer:

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Step-by-step explanation:

We notice that the answer choices have only the first equation changed. The equivalent of the first equation must have the same ratios of coefficients.

x-coefficient : y-coefficient : constant

= 2 : 1 : -15 . . . . . in the given equation

2 : -2 : 30 = 1 : -1 : 15 . . . upper left — not the same

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