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Q. 1

Expert-verifiedFound in: Page 505

Book edition
6th

Author(s)
Sullivan

Pages
1200 pages

ISBN
9780321795465

Find the real solutions, if any, of the equation $3{x}^{2}+x-1=0$.

The solutions are $\left\{\frac{-1-\sqrt{13}}{6},\frac{-1+\sqrt{13}}{6}\right\}$.

We are given the equation $3{x}^{2}+x-1=0$ and we need to find the real roots of it.

The standard form of a quadratic equation is $a{x}^{2}+bx+c=0$.

Comparing it with the given equation we get,

$a=3,b=1,c=-1$.

Discriminant of the equation is

role="math" localid="1646502307564" ${b}^{2}-4ac\phantom{\rule{0ex}{0ex}}={\left(1\right)}^{2}-4\left(3\right)(-1)\phantom{\rule{0ex}{0ex}}=1+12\phantom{\rule{0ex}{0ex}}=13>0$

Thus there are two unequal real solutions for $x$.

Using the quadratic formula to find the roots:

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

Substituting the values we get,

role="math" localid="1646502325187" $x=\frac{-1\pm \sqrt{{1}^{2}-4\left(3\right)\left(-1\right)}}{2\left(3\right)}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{-1\pm \sqrt{1+12}}{6}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{-1\pm \sqrt{13}}{6}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{-1-\sqrt{13}}{6},\frac{-1+\sqrt{13}}{6}$

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