Question

HELP URGENTLY PLEASE BRAINLIEST!!

Answer 1

Given:

Two equations-

• y= 3x+5-------------(1)
• y= 4x²+x------------(2)

To Find:

Intersection of given curves or values of x and y for which both curves intersect at a point

Solution:

On equating (1) and (2), we get

$\longrightarrow\mathrm{3x+5= 4x^{2}+x}$

$\longrightarrow\mathrm{4x^{2}+x=3x+5}$

$\longrightarrow\mathrm{4x^{2}+x-3x-5=0}$

$\longrightarrow\mathrm{4x^{2}-2x-5=0}$

$\longrightarrow\mathrm{4x^{2}-2x-5=0}$

On applying quadratic formula, we get

$\mathrm{x=\dfrac{-(-2)\pm\sqrt{(-2)^{2}-4(4)(-5)}}{2(4)}}$

$\mathrm{x=\dfrac{2\pm\sqrt{4+80}}{8}}$

$\mathrm{x=\dfrac{2\pm\sqrt{84}}{8}}$

$\mathrm{x=\dfrac{2\pm\sqrt{4(21)}}{8}}$

$\mathrm{x=\dfrac{2\pm2\sqrt{21}}{8}}$

$\mathrm{x=\dfrac{2(1\pm\sqrt{21})}{8}}$

$\mathrm{x=\dfrac{1\pm\sqrt{21}}{4}}$

$\mathrm{x=\dfrac{1+\sqrt{21}}{4},\dfrac{1-\sqrt{21}}{4}}$

We know that,

$\sqrt{2}=4.583$

So,

$\mathrm{x=\dfrac{1+4.583}{4},\dfrac{1-4.583}{4}}$

$\mathrm{x=\dfrac{5.583}{4},\dfrac{-3.583}{4}}$

$\mathrm{x=1.39,-0.89}$

On putting values of x in (1), we get

Case-1: When x=1.39

$\longrightarrow{\mathrm{y=3(1.39)+5}}$

$\longrightarrow{\mathrm{y=4.17+5}}$

$\longrightarrow\mathrm{y=9.17}$

Case-2: When x= -0.89

$\longrightarrow{\mathrm{y=3(-0.89)+5}}$

$\longrightarrow{\mathrm{y=-2.67+5}}$

$\longrightarrow\mathrm{y=2.33}$