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melomori [17]
3 years ago
5

Solve each problem and show the work for each one.

Mathematics
1 answer:
Cerrena [4.2K]3 years ago
8 0

Answer:

Solve by factoring.

1. 2x^2+9x+26=-4x+5

2x²+9x+4x+26-5=0

2x²+13x+21=0

doing middle term factorization

2x²+7x+6x+21=0

x(2x+7)+3(2x+7)=0

<u>(</u><u>2</u><u>x</u><u>+</u><u>7</u><u>)</u><u>(</u><u>x</u><u>+</u><u>3</u><u>)</u><u>=</u><u>0</u>

<u>either</u>

<u>x</u><u>=</u><u>-</u><u>7</u><u>/</u><u>2</u>

<u>or</u>

<u>x</u><u>=</u><u>-</u><u>3</u><u>.</u>

2.

3x^2-6x-4=0

comparing above equation with ax²+bx+c=0

we get

a=3

b=-6

c=-4

By using quadratic equation

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

Substituting value

x=\frac{ 6±\sqrt{ {-6}^{2} - 4*3*-4 }}{2*-4}

x=\frac{ 6±\sqrt{ {84}}}{-8}

x=\frac{ 6±2\sqrt{ {21}}}{-8}

-8x=6±2√21

taking positive

-8x=6+2√21

x=\frac{6+2√21}{-8}

x=-\frac{3+√21}{4}

taking negative

-8x=6-2√21

x=\frac{6-2√21}{-8}

x=-\frac{3-√21}{4}

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