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3 years ago

I'm lost.. How do you even start it???

Mathematics

2 answers:

Anastasy [175]3 years ago

7 0

$2x+y=4\\
-x+y=1\\\\
2x+y=4\\
\underline{x-y=-1}\\
3x=3\\
x=1\\\\
1-y=-1\\
y=2
$

Sladkaya [172]3 years ago

3 0

A) 2x + y = 4
B) -x + y = 1

Multiply B) by 2
B) -2x + 2y = 2 then add to A)
A) 2x + y = 4

3 y = 6
y = 2

B) -x + y = 1
-x +2 =1
x = 1

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Thus, the two root of the given quadratic equation $x^2+4=6x$ is 5.24 and 0.76 .

Step-by-step explanation:

Consider, the given Quadratic equation, $x^2+4=6x$

This can be written as , $x^2-6x+4=0$

We have to solve using quadratic formula,

For a given quadratic equation $ax^2+bx+c=0$ we can find roots using,

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ ...........(1)

Where, $\sqrt{b^2-4ac}$ is the discriminant.

Here, a = 1 , b = -6 , c = 4

Substitute in (1) , we get,

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$\Rightarrow x=\frac{-(-6)\pm\sqrt{(-6)^2-4\cdot 1 \cdot (4)}}{2 \cdot 1}$

$\Rightarrow x=\frac{6\pm\sqrt{20}}{2}$

$\Rightarrow x=\frac{6\pm 2\sqrt{5}}{2}$

$\Rightarrow x={3\pm \sqrt{5}}$

$\Rightarrow x_1={3+\sqrt{5}}$ and $\Rightarrow x_2={3-\sqrt{5}}$

We know $\sqrt{5}=2.23607$ (approx)

Substitute, we get,

$\Rightarrow x_1={3+2.23607}$ (approx) and $\Rightarrow x_2={3-2.23607}$ (approx)

$\Rightarrow x_1={5.23607}$ (approx) and $\Rightarrow x_2=0.76393}$ (approx)

Thus, the two root of the given quadratic equation $x^2+4=6x$ is 5.24 and 0.76 .

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3 years ago

Read 2 more answers

I dont the know the answer to c-d=b for c

Vinvika [58]

C-d=b add to both sides d and will get

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