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

Comparing the quadratic equation $5x^{2}-4x-1=0$ with the standard form $ax^{2}+bx+c=0$,we get $b=\ldots$

$A$ natural number,when increased by $12$,equals $160$ times its reciprocal. Find the number.

Solve the following equation using the quadratic formula,if the equation has a solution in $R$: $5x^{2} - 3x - 2 = 0$.

Find the discriminant of the following quadratic equation: $x^{2}-7x+10=0$

If the roots of the following quadratic equation exist,find them by the method of completing the square: $3x - \frac{2}{x} - 7 = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module