Substitution To Solve Equations
Gbogbo ọrọ náà
Some equations look frightening until you notice a hidden quadratic inside them. By naming a repeated chunk with a new letter, you can turn a quartic, or an equation in powers of \(e^x\), into a plain quadratic you already know how to solve. This change of variable is one of the most elegant tricks in algebra.
In this lesson you will learn to spot a substitution such as \(u=x^2\), solve the resulting quadratic, and then return to the original variable to finish. The classic example is \(x^4-5x^2+4=0\), which becomes \(u^2-5u+4=0\) and unlocks four solutions.
Ebumnobi
- Use a substitution to form and solve a quadratic equation in order to solve a related equation.
Akọmọ Ojú-ẹkọ
Ọ dị na ngwa Green Bridge
Budata ngwa Green Bridge CBT na ekwentị maọbụ kọmputa gị iji nweta akwụkwọ ndụmọdụ zuru oke, ajụjụ mmụta, na ndị ọzọ.
Ayẹwo Ẹkọ
Ekele diri gi maka imecha ihe karịrị na Substitution To Solve Equations. Ugbu a na ị na-enyochakwa isi echiche na echiche ndị dị mkpa, ọ bụ oge iji nwalee ihe ị ma. Ngwa a na-enye ụdị ajụjụ ọmụmụ dị iche iche emebere iji kwado nghọta gị wee nyere gị aka ịmata otú ị ghọtara ihe ndị a kụziri.
Ị ga-ahụ ngwakọta nke ụdị ajụjụ dị iche iche, gụnyere ajụjụ chọrọ ịhọrọ otu n’ime ọtụtụ azịza, ajụjụ chọrọ mkpirisi azịza, na ajụjụ ede ede. A na-arụpụta ajụjụ ọ bụla nke ọma iji nwalee akụkụ dị iche iche nke ihe ọmụma gị na nkà nke ịtụgharị uche.
Jiri akụkụ a nke nyocha ka ohere iji kụziere ihe ị matara banyere isiokwu ahụ ma chọpụta ebe ọ bụla ị nwere ike ịchọ ọmụmụ ihe ọzọ. Ekwela ka nsogbu ọ bụla ị na-eche ihu mee ka ị daa mba; kama, lee ha anya dị ka ohere maka ịzụlite onwe gị na imeziwanye.
- Using u = x^2, the equation x^4 - 5x^2 + 4 = 0 becomes: A. u^2 - 5u + 4 = 0 B. u^2 + 5u + 4 = 0 C. u - 5u + 4 = 0 D. 2u - 5 = 0 Answer: A
- How many real solutions has x^4 - 5x^2 + 4 = 0? A. 1 B. 2 C. 3 D. 4 Answer: D
- Solve x^4 - 5x^2 + 4 = 0. A. x = 1 or x = 4 B. x = +/-1 or x = +/-2 C. x = +/-1 or x = +/-4 D. x = 2 only Answer: B
- For e^{2x} - 3e^x + 2 = 0, the substitution u = e^x gives the values: A. u = 1 or u = 2 B. u = -1 or u = -2 C. u = 0 or u = 3 D. u = 2 only Answer: A
- Solve e^{2x} - 3e^x + 2 = 0. A. x = 0 or x = ln 2 B. x = 1 or x = 2 C. x = ln 2 only D. x = -1 or x = -2 Answer: A
Ọ dị na ngwa Green Bridge
Budata ngwa Green Bridge CBT na ekwentị maọbụ kọmputa gị iji nweta akwụkwọ ndụmọdụ zuru oke, ajụjụ mmụta, na ndị ọzọ.
Ọ dị na ngwa Green Bridge
Budata ngwa Green Bridge CBT na ekwentị maọbụ kọmputa gị iji nweta akwụkwọ ndụmọdụ zuru oke, ajụjụ mmụta, na ndị ọzọ.
Àwọn Ìwé Tó Dáa Láti Kà
Meecha Ajụjụ Ule Ọmarịcha
Ị chọrọ ime ajụjụ ule ọmarịcha gbasara Substitution To Solve Equations? Budata ngwa Green Bridge CBT iji nweta ajụjụ ule ọmarịcha na nyocha zuru ezu gbasara isiokwu a.
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