Question

only the final answer is also ok....


Form a quadratic equation if the roots of the quadratic equation are 2 + √7 and 2 - √7 ​

Answer 1

ANSWER:

Given:

• Zeroes of polynomial = 2 + √7, 2 - √7

To Find:

• The quadratic polynomial

Solution:

$\text{\underline{METHOD 1:}}\\\\\text{We are given that, zeroes are 2+ \sqrt7 and 2- \sqrt7 . So,}\\\\:\implies x=2+\sqrt7 \:\:\:and\:\:\:x=2-\sqrt7 \\\\:\implies x-3=0\:\:\:and\:\:\:x-5=0\\\\\text{Multiplying the equations,}\\\\:\implies(x-(2+\sqrt7))\times(x-(2-\sqrt7)) =0\times0\\\\:\implies (x-2-\sqrt7)(x-2+\sqrt7)=0\\\\:\implies x^2-2x+x\sqrt7\!\!\!\!\!/\:-2x+4-2\sqrt7\!\!\!\!\!/\:-x\sqrt7\!\!\!\!\!/\:+2\sqrt7\!\!\!\!\!/\:-7=0\\\\:\implies x^2-2x-2x+4-7=0\\\\\bf{:\implies x^2-4x-3}$

$\\\text{\underline{METHOD 2:}}\\\\\text{We know that,}\\\\:\implies\text{k[x ^2 - (sum of zeroes)x + (product of zeroes)]}\\\\\text{We are given that, zeroes are 2+ \sqrt7 and 2- \sqrt7 . So,}\\\\:\implies k[x^2-(2+\sqrt7\!\!\!\!/\:+2-\sqrt7\!\!\!\!/)x+((2+\sqrt7)\times(2-\sqrt7))]\\\\\text{We know that, a^2-b^2=(a+b)(a-b) . So,}\\\\:\implies k[x^2-4x+(2^2-(\sqrt7)^2)]\\\\:\implies k[x^2-4x+(4-7)]\\\\:\implies k[x^2-4x+(-3)]\\\\:\implies k[x^2-4x-3]\\\\\bf{:\implies x^2-4x-3}$