Find the roots of the following quadratic equ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Comparing the equation $y^{2}+10y+6=0$ with the standard form $ay^{2}+by+c=0$,we get $a=1, b=10, c=6$.The discriminant $D$ is given by $D = b^{2}-

Vedclass · Accepted Answer

$-5+\sqrt{19}$ and $-5-\sqrt{19}$

Vedclass · Answer

(B) Comparing the equation $y^{2}+10y+6=0$ with the standard form $ay^{2}+by+c=0$,we get $a=1, b=10, c=6$.The discriminant $D$ is given by $D = b^{2}-4ac$.$D = (10)^{2} - 4(1)(6) = 100 - 24 = 76$.Since $D > 0$,the equation has two distinct real roots.The quadratic formula is $y = \frac{-b \pm \sqrt{D}}{2a}$.Substituting the values,we get $y = \frac{-10 \pm \sqrt{76}}{2(1)}$.Since $\sqrt{76} = \sqrt{4 	imes 19} = 2\sqrt{19}$,we have $y = \frac{-10 \pm 2\sqrt{19}}{2}$.Dividing by $2$,we get $y = -5 \pm \sqrt{19}$.Thus,the roots are $-5+\sqrt{19}$ and $-5-\sqrt{19}$.

Find the roots of the following quadratic equation by using the quadratic formula,if they exist: $y^{2}+10y+6=0$

Similar Questions

What is the value of $k$ if one of the roots of the quadratic equation $x^{2} + 6x + k = 0$ is $-4$?

Check whether the following is a quadratic equation:
$(2x - 1)(x - 3) = (x + 5)(x - 1)$

Find the roots of the following quadratic equation by the method of completing the square: $x^{2}-8x+12=0$

Verify whether the given value of $x$ is a solution of the quadratic equation or not: $2x^{2} - 6x + 3 = 0$; $x = \frac{1}{2}$.

In Hindu civilization,$\ldots \ldots \ldots$ gave the method to solve quadratic equations.

Vedclass Test Series

Exam Paper Generator

Online Exam Module