Question

निम्नांकित द्विघात समीकरण का आदित्य अद्वितीय हल बज्र –गुणन विधि से ज्ञात कीजिए – 2x+y = 5 3x+2y= 8

Answer 1

Answer

I don't understand

Step-by-step explanation:

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Answer 2

The quadratic equation is $-6x^2+2y^2-xy+x-18y+40=0$

Step-by-step explanation:

Given equations are -2x+y=5 and 3x+2y=8

Now to find the quadratic equations with these equations by solving equations in multiplication method

-2x+y-5=0 and 3x+2y-8=0

Now multiply the two equations

• (-2x+y-5)(3x+2y-8)=0
• (-2x)(3x+2y-8)+y(3x+2y-8)-5(3x+2y-8)=0 ( using the distributive property a(x+y+z)=ax+ay+az )
• (-2x)(3x)+(-2x)(2y)+(-2x)(-8)+(y)(3x)+y(2y)+y(-8)+(-5)(3x)+(-5)(2y)+(-5)(-8)=0
• $-6x^2-4xy+16x+3xy+2y^2-8y-15x-10y+40=0$
• $-6x^2+2y^2-xy+x-18y+40=0$

The quadratic equation is $-6x^2+2y^2-xy+x-18y+40=0$